Thermal Physics Alıştırma Soruları — JAMB UTME — Sayfa 2

Soru 21Soru →

A flexible balloon contains a sample of gas occupying a volume of $3.0\text{ m}^3$ at a temperature of $27^\circ\text{C}$ under constant pressure. If the gas is heated to $127^\circ\text{C}$ while keeping the pressure constant, what is the new volume of the gas in $\text{m}^3$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The new volume of the gas is

4.0\text{ m}^3

.

According to Charles's Law (

\frac{V_1}{T_1} = \frac{V_2}{T_2}

), at constant pressure, volume is directly proportional to absolute temperature in Kelvin. Converting

27^\circ\text{C}

to

300\text{ K}

and

127^\circ\text{C}

to

400\text{ K}

yields a final volume

V_2 = 3.0 \times \frac{400}{300} = 4.0\text{ m}^3

.

Adım Adım Çözüm

1

Convert the given initial and final temperatures from Celsius to Kelvin

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 127 + 273 = 400\text{ K}

Gas law calculations require absolute temperature in Kelvin.

2

Set up Charles's Law equation for constant pressure processes

\frac{V_1}{T_1} = \frac{V_2}{T_2}

At constant pressure, the volume of a given mass of gas is directly proportional to its absolute temperature.

3

Substitute the values into the equation and solve for the final volume

V_2

V_2 = 3.0 \times \frac{400\text{ K}}{300\text{ K}} = 4.0\text{ m}^3

Direct algebraic evaluation yields the final gas volume.

Anahtar Kavram

Charles's Law

Soru 22Soru →

A meteorological balloon filled with $0.50\text{ kg}$ of helium gas is launched at sea level, where the atmospheric pressure is $1.01 \times 10^5\text{ Pa}$ and the ambient temperature is $27^\circ\text{C}$ . The balloon ascends to a high altitude where the external ambient pressure decreases to $4.00 \times 10^4\text{ Pa}$ and the ambient temperature drops to $-23^\circ\text{C}$ . As the balloon expands, its elastic membrane exerts an additional pressure, causing the internal gas pressure to be $20\%$ higher than the surrounding ambient pressure. Assuming helium behaves as an ideal gas with a molar mass of $4.0\text{ g/mol}$ and the molar gas constant $R = 8.31\text{ J mol}^{-1}\text{K}^{-1}$ , calculate the final volume of helium gas inside the balloon at this altitude in $\text{m}^3$ .

Cevabı ve açıklamayı göster

Cevap: 5.41

Cevap

The final volume of helium gas inside the balloon at altitude is

5.41\text{ m}^3

.

The ideal gas equation

PV = nRT

relates state variables. By determining

n = 125\text{ moles}

from mass and molar mass, absolute temperature

T_2 = 250\text{ K}

, and total internal pressure

P_2 = 4.80 \times 10^4\text{ Pa}

, the final volume

V_2

evaluates to

5.41\text{ m}^3

.

Adım Adım Çözüm

1

Convert the mass of helium into moles

n = 125\text{ moles}

Mass

m = 0.50\text{ kg} = 500\text{ g}

divided by molar mass

M = 4.0\text{ g/mol}

gives

n = \frac{500}{4.0} = 125\text{ moles}

.

2

Convert final temperature to Kelvin

T_2 = 250\text{ K}

Gas laws strictly require thermodynamic temperature:

T_2 = -23 + 273 = 250\text{ K}

.

3

Calculate final internal pressure of the gas

P_2 = 4.80 \times 10^4\text{ Pa}

The gas pressure inside the balloon is

20\%

higher than external ambient pressure:

P_2 = 1.20 \times (4.00 \times 10^4) = 4.80 \times 10^4\text{ Pa}

.

4

Solve for the final volume using the ideal gas equation

V_2 = 5.41\text{ m}^3

Rearranging

P_2 V_2 = n R T_2

yields

V_2 = \frac{n R T_2}{P_2} = \frac{125 \times 8.31 \times 250}{4.80 \times 10^4} = 5.41015...\text{ m}^3 \approx 5.41\text{ m}^3

.

Anahtar Kavram

Ideal Gas Equation (

PV = nRT

)

Tahmini Süre:3m 0s

Soru 23Soru →

According to the kinetic theory of matter, what causes the pressure exerted by a gas on the walls of its container?

The continuous elastic collisions of gas molecules with the walls of the container

The attractive forces pulling gas molecules towards the container walls

The heat capacity of individual gas molecules transferring thermal mass to the walls

The static expansion of fixed gas molecules taking up space

Cevabı ve açıklamayı göster

Cevap: The continuous elastic collisions of gas molecules with the walls of the container

Cevap

The pressure exerted by a gas is due to the continuous elastic collisions of gas molecules with the container walls.

According to the kinetic theory of gases, gas particles are in rapid, random motion. When these particles collide elastically with the walls of the container, they undergo a change in momentum. The average rate of momentum change per unit area exerted by countless particle impacts manifests as macroscopic gas pressure.

Adım Adım Çözüm

1

Identify the basic postulate of the kinetic theory of gases regarding particle motion.

Gas molecules are in rapid, constant, and random motion.

Kinetic theory models gas behavior based on particle movement.

2

Relate particle motion to force and pressure on the container boundary.

When particles hit the container wall, they undergo a change in momentum, imparting a force on the wall.

Force is defined as the rate of change of momentum (

F = \frac{\Delta p}{\Delta t}

).

3

Define pressure in terms of force per unit surface area.

The total force exerted per unit surface area by continuous collisions results in gas pressure (

P = \frac{F}{A}

).

Pressure is force distributed over a given surface area.

Anahtar Kavram

Kinetic Theory of Matter and Pressure of Gases

Soru 24Soru →

A rectangular metal sheet has an initial area of $2.0\text{ m}^2$ at $20^\circ\text{C}$ . If the linear expansivity of the metal is $2.5 \times 10^{-5}\text{ K}^{-1}$ , what is the final temperature required for its area to increase by $0.005\text{ m}^2$ ?

70^\circ\text{C}

120^\circ\text{C}

50^\circ\text{C}

53.3^\circ\text{C}

Cevabı ve açıklamayı göster

Cevap: $70^\circ\text{C}$

Cevap

The final temperature required is

70^\circ\text{C}

.

Area expansion is governed by

\Delta A = A_0 \beta \Delta T

, where the area expansivity

\beta = 2\alpha

. Substituting

A_0 = 2.0\text{ m}^2

,

\Delta A = 0.005\text{ m}^2

, and

\beta = 5.0 \times 10^{-5}\text{ K}^{-1}

gives a temperature rise

\Delta T = 50^\circ\text{C}

. Adding the initial temperature of

20^\circ\text{C}

yields a final temperature of

70^\circ\text{C}

.

Adım Adım Çözüm

1

Determine the area expansivity (superficial expansivity)

\beta

from the linear expansivity

\alpha

.

\beta = 2\alpha = 2 \times 2.5 \times 10^{-5}\text{ K}^{-1} = 5.0 \times 10^{-5}\text{ K}^{-1}

Area expansion depends on area expansivity, which is twice the linear expansivity.

2

Calculate the temperature change

\Delta T

using the formula

\Delta A = A_0 \beta \Delta T

.

$\Delta T = \frac{\Delta A}{A_0 \beta} = \frac{0.005\text{ m}^2}{2.0\text{ m}^2 \times 5.0 \times 10^{-5}\text{ K}^{-1}} = \frac{5 \times 10^{-3}}{1.0 \times 10^{-4}} = 50\text{ K}$

Rearranging the expansion formula isolates the temperature change variable.

3

Calculate the final temperature

T_2

by adding

\Delta T

to the initial temperature

T_1

.

T_2 = T_1 + \Delta T = 20^\circ\text{C} + 50^\circ\text{C} = 70^\circ\text{C}

The final temperature is the sum of the initial temperature and the rise in temperature.

Anahtar Kavram

Thermal Expansion of Solids (Area Expansivity)

Tahmini Süre:1m 30s

Soru 25Soru →

A high-pressure storage vessel contains a sample of gas at an initial pressure of $2.00 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . A relief valve releases one-third of the total mass of the gas while maintaining a constant internal volume. After the valve closes, the vessel and remaining gas are heated to $127^\circ\text{C}$ . What is the final pressure of the gas inside the vessel?

1.78 \times 10^5\text{ Pa}

6.27 \times 10^5\text{ Pa}

8.89 \times 10^4\text{ Pa}

2.67 \times 10^5\text{ Pa}

Cevabı ve açıklamayı göster

Cevap: $1.78 \times 10^5\text{ Pa}$

Cevap

The final pressure of the gas inside the vessel is

1.78 \times 10^5\text{ Pa}

.

According to the ideal gas equation

PV = \frac{m}{M}RT

, pressure is directly proportional to mass and absolute temperature at fixed volume (

P \propto m T

). Converting temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Since one-third of the gas escaped, two-thirds remains (

m_2 = \frac{2}{3}m_1

). The final pressure is therefore

P_2 = P_1 \times \frac{2}{3} \times \frac{400}{300} = 2.00 \times 10^5\text{ Pa} \times \frac{8}{9} \approx 1.78 \times 10^5\text{ Pa}

.

Adım Adım Çözüm

1

Convert given initial and final temperatures from Celsius to Kelvin.

T_1 = 27^\circ\text{C} + 273 = 300\text{ K}

and

T_2 = 127^\circ\text{C} + 273 = 400\text{ K}

.

Absolute temperature in Kelvin is required for all gas law calculations.

2

Determine the remaining mass fraction of the gas.

m_2 = m_1 - \frac{1}{3}m_1 = \frac{2}{3}m_1

.

The pressure depends on the quantity of gas remaining in the rigid container after venting.

3

Apply the ideal gas equation

PV = nRT = \frac{m}{M}RT

for constant volume

V

and molar mass

M

.

\frac{P_2}{P_1} = \left(\frac{m_2}{m_1}\right) \left(\frac{T_2}{T_1}\right)

.

Pressure is directly proportional to both mass and absolute temperature when volume is constant.

4

Substitute the known values to compute

P_2

.

P_2 = (2.00 \times 10^5\text{ Pa}) \times \left(\frac{2}{3}\right) \times \left(\frac{400\text{ K}}{300\text{ K}}\right) = 2.00 \times 10^5 \times \frac{8}{9} \approx 1.78 \times 10^5\text{ Pa}

.

Evaluates the combined effects of mass reduction and temperature elevation.

Anahtar Kavram

Ideal Gas Law variations involving changing gas mass and absolute temperature at constant volume.

Soru 26Soru →

A metal block of mass $2.5\text{ kg}$ absorbs $1200\text{ J}$ of thermal energy, causing its temperature to rise by $15\text{ K}$ . What is the heat capacity of the block in $\text{J K}^{-1}$ ?

Cevabı ve açıklamayı göster

Cevap: 80

Cevap

The heat capacity of the block is

80\text{ J K}^{-1}

.

Heat capacity

C

represents the quantity of heat energy required to raise the temperature of the entire object by

1\text{ K}

. Using the relation

C = \frac{Q}{\Delta T}

, substituting

Q = 1200\text{ J}

and

\Delta T = 15\text{ K}

gives

C = \frac{1200}{15} = 80\text{ J K}^{-1}

.

Adım Adım Çözüm

1

Identify the given quantities from the problem.

Heat energy

Q = 1200\text{ J}

, temperature change

\Delta T = 15\text{ K}

, and mass

m = 2.5\text{ kg}

.

Clear identification of parameters is necessary to choose the correct formula.

2

Apply the formula for heat capacity.

C = \frac{Q}{\Delta T}

Heat capacity

C

measures the heat required to change the temperature of the entire body by

1\text{ K}

, regardless of mass per unit quantity.

3

Substitute the values to calculate the heat capacity.

C = \frac{1200}{15} = 80\text{ J K}^{-1}

Dividing total thermal energy absorbed by the resulting temperature rise yields the heat capacity.

Anahtar Kavram

Heat Capacity (

C = \frac{Q}{\Delta T}

)

Tahmini Süre:45s

Soru 27Soru →

A metallic container with an initial volume of $400 \text{ cm}^3$ at $15^\circ\text{C}$ is filled completely with paraffin. Upon heating the container and its contents to $65^\circ\text{C}$ , a volume of $18 \text{ cm}^3$ of paraffin spills over. Given that the linear expansivity of the metal container is $2.0 \times 10^{-5} \text{ K}^{-1}$ , what is the real cubic expansivity of the paraffin, expressed in units of $10^{-4} \text{ K}^{-1}$ ?

Cevabı ve açıklamayı göster

Cevap: 9.6

Cevap

The real cubic expansivity of the paraffin is

9.6 \times 10^{-4} \text{ K}^{-1}

, which corresponds to a numerical value of

9.6

in units of

10^{-4} \text{ K}^{-1}

.

The real cubic expansivity of a liquid accounts for both the observed (apparent) expansion of the liquid and the expansion of the container holding it. By applying

\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = 9.0 \times 10^{-4} \text{ K}^{-1}

and

\gamma_v = 3\alpha = 0.6 \times 10^{-4} \text{ K}^{-1}

, we sum them to obtain the real cubic expansivity

\gamma_r = 9.6 \times 10^{-4} \text{ K}^{-1}

.

Adım Adım Çözüm

1

Determine the temperature change (

\Delta T

) and the apparent change in volume (

\Delta V_a

).

\Delta T = 65^\circ\text{C} - 15^\circ\text{C} = 50 \text{ K}

and

\Delta V_a = 18 \text{ cm}^3

.

The overflow volume represents the apparent expansion of the liquid relative to the expanding container over the temperature rise.

2

Calculate the apparent cubic expansivity (

\gamma_a

) of the paraffin.

\gamma_a = \frac{\Delta V_a}{V_0 \Delta T} = \frac{18}{400 \times 50} = \frac{18}{20000} = 9.0 \times 10^{-4} \text{ K}^{-1}

.

Apparent expansivity relates the apparent volume expansion to the original volume and temperature increase.

3

Calculate the cubic expansivity of the metallic vessel (

\gamma_v

).

\gamma_v = 3 \alpha = 3 \times 2.0 \times 10^{-5} \text{ K}^{-1} = 6.0 \times 10^{-5} \text{ K}^{-1} = 0.6 \times 10^{-4} \text{ K}^{-1}

.

Cubic expansivity of an isotropic solid container is three times its linear expansivity.

4

Calculate the real cubic expansivity of the paraffin (

\gamma_r

).

\gamma_r = \gamma_a + \gamma_v = 9.0 \times 10^{-4} \text{ K}^{-1} + 0.6 \times 10^{-4} \text{ K}^{-1} = 9.6 \times 10^{-4} \text{ K}^{-1}

.

The real expansion of a liquid is the sum of its apparent expansion and the expansion of the containing vessel.

Anahtar Kavram

Real vs. Apparent Expansion of Liquids (

\gamma_r = \gamma_a + \gamma_v

where

\gamma_v = 3\alpha

)

Soru 28Soru →

A rigid vessel of fixed volume contains an ideal gas at an initial pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . Additional gas is pumped into the vessel until the total number of moles of gas is doubled. If the temperature of the gas increases to $87^\circ\text{C}$ , what is the final pressure of the gas in units of $10^5\text{ Pa}$ ?

Cevabı ve açıklamayı göster

Cevap: 3.6

Cevap

The final pressure of the gas is

3.6 \times 10^5\text{ Pa}

(which is

3.6

in units of

10^5\text{ Pa}

).

Using the ideal gas equation

PV = nRT

at fixed volume, the ratio of final to initial pressure is given by

P_2/P_1 = (n_2/n_1) \times (T_2/T_1)

. Converting temperatures to Kelvin yields

T_1 = 300\text{ K}

and

T_2 = 360\text{ K}

. Given that the number of moles doubles (

n_2/n_1 = 2

), substituting the values gives

P_2 = 1.50 \times 10^5 \times 2 \times (360/300) = 3.60 \times 10^5\text{ Pa}

.

Adım Adım Çözüm

1

Convert initial and final temperatures from Celsius to absolute temperature in Kelvin.

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 87 + 273 = 360\text{ K}

.

Gas laws require absolute temperatures in Kelvin for thermodynamic calculations.

2

Formulate the pressure relation from the ideal gas equation

PV = nRT

.

Since volume

V

and the universal gas constant

R

are constant,

P_2 = P_1 \times \frac{n_2}{n_1} \times \frac{T_2}{T_1}

.

Pressure is directly proportional to both the number of moles and the absolute temperature when volume is fixed.

3

Substitute the mole ratio

n_2/n_1 = 2

, initial pressure, and Kelvin temperatures to compute

P_2

.

P_2 = 1.50 \times 10^5 \times 2 \times \frac{360}{300} = 3.60 \times 10^5\text{ Pa}

.

Evaluates the final gas pressure in the requested numerical units.

Anahtar Kavram

Ideal Gas Equation and Variable Moles under Constant Volume

Soru 29Soru →

An electric heater rated at $690\text{ W}$ is immersed in a thermally insulated container holding a mixture of $0.50\text{ kg}$ of ice and $0.50\text{ kg}$ of liquid water in equilibrium at $0^\circ\text{C}$ . The heater is operated for $5.0\text{ minutes}$ . Assuming negligible heat capacity for the container and no heat loss to the surroundings, what is the final equilibrium temperature of the mixture?
(Specific latent heat of fusion of ice $L_f = 3.3 \times 10^5\text{ J kg}^{-1}$ , specific heat capacity of water $c_w = 4200\text{ J kg}^{-1}\text{ K}^{-1}$ )

0^\circ\text{C}

10^\circ\text{C}

20^\circ\text{C}

5^\circ\text{C}

Cevabı ve açıklamayı göster

Cevap: $10^\circ\text{C}$

Cevap

The final equilibrium temperature of the mixture is

10^\circ\text{C}

.

The total energy supplied by the

690\text{ W}

heater in

300\text{ s}

is

207,000\text{ J}

. Melting all

0.50\text{ kg}

of ice at

0^\circ\text{C}

requires

165,000\text{ J}

. The remaining

42,000\text{ J}

heats the combined

1.00\text{ kg}

of water (initial water + melted ice) through

\Delta T = \frac{42,000}{1.00 \times 4200} = 10^\circ\text{C}

, reaching a final equilibrium temperature of

10^\circ\text{C}

.

Adım Adım Çözüm

1

Calculate the total thermal energy provided by the electric heater.

Q_{\text{total}} = P \times t = 690\text{ W} \times (5.0 \times 60\text{ s}) = 207,000\text{ J}

.

Power multiplied by time yields total energy supplied.

2

Determine the energy required to completely melt the

0.50\text{ kg}

of ice at

0^\circ\text{C}

.

Q_{\text{melt}} = m_{\text{ice}} \times L_f = 0.50\text{ kg} \times 330,000\text{ J kg}^{-1} = 165,000\text{ J}

.

Latent heat of fusion changes state from solid ice to liquid water at constant temperature

0^\circ\text{C}

.

3

Calculate the remaining thermal energy available to increase the temperature.

Q_{\text{rem}} = 207,000\text{ J} - 165,000\text{ J} = 42,000\text{ J}

.

After complete melting, excess energy goes into sensible heating.

4

Calculate the total mass of liquid water and the resulting temperature rise.

m_{\text{total}} = 0.50\text{ kg (melted ice)} + 0.50\text{ kg (initial water)} = 1.00\text{ kg}

.

\Delta T = \frac{Q_{\text{rem}}}{m_{\text{total}} \times c_w} = \frac{42,000\text{ J}}{1.00\text{ kg} \times 4200\text{ J kg}^{-1}\text{ K}^{-1}} = 10^\circ\text{C}

.

All water now absorbs energy to raise the temperature.

Anahtar Kavram

Phase changes occur at constant temperature (latent heat

Q = mL

). Once the phase change is complete, additional thermal energy increases temperature as sensible heat (

Q = mc\Delta T

) using the total combined mass.

Soru 30Soru →

A uniform metal rod of length $0.50\text{ m}$ and cross-sectional area $2.0 \times 10^{-3}\text{ m}^2$ has a thermal conductivity of $400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . If a temperature difference of $50\text{ K}$ is maintained between its ends, what is the rate of heat flow through the rod?

80\text{ W}

40\text{ W}

20\text{ W}

160\text{ W}

Cevabı ve açıklamayı göster

Cevap: $80\text{ W}$

Cevap

The rate of heat flow through the rod is

80\text{ W}

.

The rate of heat transfer by conduction is calculated using Fourier's law,

\frac{Q}{t} = \frac{k A \Delta T}{d}

. Substituting

k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

,

A = 2.0 \times 10^{-3}\text{ m}^2

,

\Delta T = 50\text{ K}

, and

d = 0.50\text{ m}

yields

80\text{ W}

.

Adım Adım Çözüm

1

Identify the given parameters for thermal conduction.

Thermal conductivity

k = 400\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

, cross-sectional area

A = 2.0 \times 10^{-3}\text{ m}^2

, temperature difference

\Delta T = 50\text{ K}

, and length

d = 0.50\text{ m}

.

These parameters form the components of Fourier's law of thermal conduction.

2

Apply the conduction rate formula

\frac{Q}{t} = \frac{k A \Delta T}{d}

.

\frac{Q}{t} = \frac{400 \times (2.0 \times 10^{-3}) \times 50}{0.50}

Heat transfer per unit time depends directly on conductivity, area, and temperature difference, and inversely on thickness/length.

3

Calculate the numerical value.

\frac{Q}{t} = \frac{40}{0.50} = 80\text{ W}

Dividing the product of the numerator terms (

40\text{ J/s}\cdot\text{m}

) by length (

0.50\text{ m}

) yields the rate of energy transfer.

Anahtar Kavram

Thermal Conduction Rate (Fourier's Law)

Tahmini Süre:45s

Soru 31Soru →

A sample of gas enclosed in a rigid container exerts a pressure of $3.0 \times 10^5 \text{ N m}^{-2}$ with a density of $0.40 \text{ kg m}^{-3}$ . If the gas is heated at constant volume until its pressure increases to $1.2 \times 10^6 \text{ N m}^{-2}$ , what is the final root-mean-square (r.m.s.) speed of the gas molecules?

3000 \text{ m s}^{-1}

6000 \text{ m s}^{-1}

1500 \text{ m s}^{-1}

750 \text{ m s}^{-1}

Cevabı ve açıklamayı göster

Cevap: $3000 \text{ m s}^{-1}$

Cevap

The final root-mean-square speed of the gas molecules is

3000 \text{ m s}^{-1}

.

Using the kinetic theory equation

P = \frac{1}{3}\rho v_{\text{rms}}^2

, the initial speed is

v_1 = \sqrt{\frac{3 \times 3.0 \times 10^5}{0.40}} = 1500 \text{ m s}^{-1}

. When heated at constant volume, density remains unchanged. The new pressure

1.2 \times 10^6 \text{ N m}^{-2}

is 4 times the initial pressure, so the new speed is

v_2 = \sqrt{4} \times v_1 = 2 \times 1500 \text{ m s}^{-1} = 3000 \text{ m s}^{-1}

.

Adım Adım Çözüm

1

Express the relationship between pressure, density, and root-mean-square speed using kinetic theory.

P = \frac{1}{3}\rho v_{\text{rms}}^2 \implies v_{\text{rms}} = \sqrt{\frac{3P}{\rho}}

According to the kinetic theory of gases, the pressure exerted by a gas is related to its density and molecular r.m.s. speed.

2

Calculate the initial root-mean-square speed

v_1

using initial pressure

P_1 = 3.0 \times 10^5 \text{ N m}^{-2}

and density

\rho = 0.40 \text{ kg m}^{-3}

.

v_1 = \sqrt{\frac{3 \times (3.0 \times 10^5)}{0.40}} = \sqrt{\frac{9.0 \times 10^5}{0.40}} = \sqrt{2.25 \times 10^6} = 1500 \text{ m s}^{-1}

Establishes the baseline molecular speed prior to heating.

3

Determine the final root-mean-square speed

v_2

after the pressure increases to

P_2 = 1.2 \times 10^6 \text{ N m}^{-2}

at constant volume.

v_2 = \sqrt{\frac{3 \times (1.2 \times 10^6)}{0.40}} = \sqrt{\frac{3.6 \times 10^6}{0.40}} = \sqrt{9.0 \times 10^6} = 3000 \text{ m s}^{-1}

Since the volume is rigid, density

\rho

remains constant while pressure increases due to heating.

Anahtar Kavram

Root-Mean-Square Speed and Pressure Relation in Kinetic Theory

Soru 32Soru →

A thermometer calibrated on an arbitrary scale $X$ registers a lower fixed point of $-10^\circ\text{X}$ and an upper fixed point of $110^\circ\text{X}$ . What is the true temperature in degrees Celsius ( $^\circ\text{C}$ ) when this thermometer reads $20^\circ\text{X}$ ?

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

The true temperature on the Celsius scale is

25^\circ\text{C}

.

Using the relation

\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta}{100}

, substituting

X = 20

,

\text{LFP}_X = -10

, and

\text{UFP}_X = 110

gives

\frac{20 - (-10)}{110 - (-10)} = \frac{30}{120} = 0.25

. Multiplying

0.25

by

100

gives

25^\circ\text{C}

.

Adım Adım Çözüm

1

Set up the linear relationship between the arbitrary temperature scale

X

and the Celsius scale

\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta - \text{LFP}_C}{\text{UFP}_C - \text{LFP}_C}

Thermometric properties vary linearly with temperature between fixed points.

2

Substitute the given numerical values into the formula

\frac{20 - (-10)}{110 - (-10)} = \frac{\theta - 0}{100 - 0} \implies \frac{30}{120} = \frac{\theta}{100}

The lower fixed point on scale

X

is

-10^\circ\text{X}

and the upper fixed point is

110^\circ\text{X}

.

3

Solve for the unknown temperature

\theta

in degrees Celsius

\theta = \frac{30}{120} \times 100 = 25^\circ\text{C}

Simplifying the fraction

\frac{30}{120}

yields

\frac{1}{4}

, and

\frac{1}{4} \times 100 = 25

.

Anahtar Kavram

Linear interpolation and conversion between thermometric temperature scales

Soru 33Soru →

An electric immersion heater rated at $200\text{ W}$ is used to heat $0.8\text{ kg}$ of a liquid contained in a vessel of heat capacity $120\text{ J K}^{-1}$ . The initial temperature of the liquid and vessel is $20^\circ\text{C}$ . If the heater is operated for $3.5\text{ minutes}$ and the final temperature reaches $50^\circ\text{C}$ , calculate the specific heat capacity of the liquid in $\text{J kg}^{-1}\text{ K}^{-1}$ , assuming no thermal energy is lost to the surroundings.

Cevabı ve açıklamayı göster

Cevap: 1600

Cevap

The specific heat capacity of the liquid is

1600\text{ J kg}^{-1}\text{ K}^{-1}

.

Total energy supplied by the heater is

Q = 200\text{ W} \times 210\text{ s} = 42,000\text{ J}

. The temperature increase is

\Delta T = 30\text{ K}

. The energy absorbed by the container is

Q_{\text{vessel}} = C \Delta T = 120 \times 30 = 3,600\text{ J}

. The remaining energy

38,400\text{ J}

is absorbed by the liquid. Dividing this value by the product of the mass of liquid (

0.8\text{ kg}

) and temperature rise (

30\text{ K}

) gives the specific heat capacity

1600\text{ J kg}^{-1}\text{ K}^{-1}

.

Adım Adım Çözüm

1

Convert the heating time to seconds and compute total heat energy supplied by the heater.

Q = P \times t = 200\text{ W} \times (3.5 \times 60\text{ s}) = 42,000\text{ J}

Heat energy supplied by an electric source equals electrical power multiplied by duration in seconds.

2

Determine the change in temperature of the system.

\Delta T = 50^\circ\text{C} - 20^\circ\text{C} = 30\text{ K}

Both the vessel and liquid experience the same initial and final temperatures.

3

Calculate the heat energy absorbed by the vessel.

Q_{\text{vessel}} = C \times \Delta T = 120\text{ J K}^{-1} \times 30\text{ K} = 3,600\text{ J}

Heat capacity

C

represents heat required per unit temperature rise for the container as a whole.

4

Subtract vessel absorption from total heat supplied to find heat absorbed by the liquid.

Q_{\text{liquid}} = 42,000\text{ J} - 3,600\text{ J} = 38,400\text{ J}

Conservation of energy dictates

Q_{\text{total}} = Q_{\text{vessel}} + Q_{\text{liquid}}

.

5

Calculate the specific heat capacity

c

of the liquid.

c = \frac{Q_{\text{liquid}}}{m \Delta T} = \frac{38,400}{0.8 \times 30} = 1600\text{ J kg}^{-1}\text{ K}^{-1}

Specific heat capacity isolates heat absorbed per unit mass per Kelvin.

Anahtar Kavram

Principle of conservation of thermal energy combining heat capacity of a vessel (

C

) and specific heat capacity of a liquid (

c

).

Tahmini Süre:1m 30s

Soru 34Soru →

A vessel of negligible heat capacity contains $0.20\text{ kg}$ of a liquid at its boiling point of $100^\circ\text{C}$ . An electric heating element rated at $500\text{ W}$ is immersed in the liquid and switched on for $3.0\text{ minutes}$ . If $0.040\text{ kg}$ of the liquid is vaporized during this period, what is the specific latent heat of vaporization of the liquid?

2.25 \times 10^6\text{ J kg}^{-1}

2.25 \times 10^4\text{ J kg}^{-1}

4.50 \times 10^5\text{ J kg}^{-1}

1.25 \times 10^4\text{ J kg}^{-1}

Cevabı ve açıklamayı göster

Cevap: $2.25 \times 10^6\text{ J kg}^{-1}$

Cevap

The specific latent heat of vaporization of the liquid is

2.25 \times 10^6\text{ J kg}^{-1}

.

The electrical energy supplied by the heater in

180\text{ seconds}

is

Q = 500 \times 180 = 90,000\text{ J}

. Since phase change occurs at constant boiling temperature (

100^\circ\text{C}

), the heat required to vaporize

0.040\text{ kg}

of liquid is given by

Q = m L_v

. Solving for

L_v

yields

\frac{90,000\text{ J}}{0.040\text{ kg}} = 2.25 \times 10^6\text{ J kg}^{-1}

.

Adım Adım Çözüm

1

Convert the time duration into seconds

t = 3.0\text{ minutes} \times 60\text{ s/min} = 180\text{ s}

SI units require time to be in seconds when calculating electrical energy.

2

Calculate total electrical thermal energy supplied by the heater

Q = P \times t = 500\text{ W} \times 180\text{ s} = 90,000\text{ J}

Thermal energy produced equals power multiplied by time elapsed.

3

Apply the latent heat formula using the mass of liquid vaporized

L_v = \frac{Q}{m} = \frac{90,000\text{ J}}{0.040\text{ kg}} = 2,250,000\text{ J kg}^{-1} = 2.25 \times 10^6\text{ J kg}^{-1}

During vaporization at constant boiling temperature, heat energy supplied goes entirely into changing state.

Anahtar Kavram

Specific Latent Heat of Vaporization

Soru 35Soru →

When air containing unsaturated water vapour is cooled at constant atmospheric pressure without adding or removing moisture, the saturated vapour pressure of water decreases while the actual partial vapour pressure of water remains constant until the dew point is reached.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

True. Saturated vapour pressure decreases as temperature falls, while actual partial vapour pressure remains constant under constant total pressure until condensation starts at the dew point.

The statement is correct because saturated vapour pressure is a temperature-dependent property that decreases as air cools. So long as total pressure remains unchanged and no water vapour is added or removed, the actual partial vapour pressure of the water vapour stays constant until the dew point is reached, at which point the air becomes saturated (

100\%

relative humidity) and condensation begins.

Adım Adım Çözüm

1

Analyze the temperature dependence of saturated vapour pressure (SVP).

SVP decreases as temperature drops.

SVP is determined by the kinetic energy of water molecules escaping into vapour at dynamic equilibrium, which decreases with decreasing temperature.

2

Determine the behavior of the actual partial vapour pressure of water during cooling at constant total pressure.

Actual partial vapour pressure remains constant prior to condensation.

Dalton's law dictates that partial pressure depends on the mole fraction of water vapour and total pressure; since no water vapour is added or removed, actual partial vapour pressure does not change.

3

Evaluate the condition for reaching the dew point.

At the dew point, SVP drops to equal the constant actual partial vapour pressure, achieving 100% relative humidity.

Relative humidity is defined as

\text{R.H.} = \frac{P_{\text{actual}}}{P_{\text{SVP}}} \times 100\%

; as

P_{\text{SVP}}

decreases towards

P_{\text{actual}}

,

\text{R.H.}

increases to

100\%

.

Anahtar Kavram

Saturated vs. actual vapour pressure, temperature dependence of SVP, and dew point

Soru 36Soru →

A composite furnace wall consists of two tightly joined layers of equal cross-sectional area. Layer 1 has a thickness of $0.02\text{ m}$ and thermal conductivity of $200\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . Layer 2 has a thickness of $0.04\text{ m}$ and thermal conductivity of $100\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . The outer surface of Layer 1 is maintained at $120^\circ\text{C}$ and the outer surface of Layer 2 is maintained at $20^\circ\text{C}$ . Under steady-state conditions, what is the rate of heat transfer per unit area through the composite wall?

200\text{ kW}\cdot\text{m}^{-2}

250\text{ kW}\cdot\text{m}^{-2}

500\text{ kW}\cdot\text{m}^{-2}

100\text{ kW}\cdot\text{m}^{-2}

Cevabı ve açıklamayı göster

Cevap: $200\text{ kW}\cdot\text{m}^{-2}$

Cevap

The rate of heat transfer per unit area through the composite furnace wall is

200\text{ kW}\cdot\text{m}^{-2}

.

Under steady-state conduction through composite layers in series, the total thermal resistance per unit area is the sum of the individual thermal resistances:

R_{\text{total}} = \frac{d_1}{k_1} + \frac{d_2}{k_2} = 1.0 \times 10^{-4} + 4.0 \times 10^{-4} = 5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

. Applying Fourier's law of heat conduction, the heat flux is

\frac{\Delta T}{R_{\text{total}}} = \frac{100}{5.0 \times 10^{-4}} = 200,000\text{ W}\cdot\text{m}^{-2} = 200\text{ kW}\cdot\text{m}^{-2}

.

Adım Adım Çözüm

1

Calculate the thermal resistance per unit area (

R

) for each layer using

R_i = \frac{d_i}{k_i}

.

R_1 = \frac{0.02\text{ m}}{200\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}} = 1.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

and

R_2 = \frac{0.04\text{ m}}{100\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}} = 4.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

.

Thermal resistance quantifies a material's opposition to conductive heat flow per unit cross-sectional area.

2

Find the total thermal resistance per unit area (

R_{\text{total}}

) by summing the series resistances.

R_{\text{total}} = R_1 + R_2 = (1.0 \times 10^{-4}) + (4.0 \times 10^{-4}) = 5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

.

Layers in series add their thermal resistances directly.

3

Determine the rate of heat flow per unit area

\frac{Q/t}{A} = \frac{\Delta T}{R_{\text{total}}}

.

\frac{Q/t}{A} = \frac{120^\circ\text{C} - 20^\circ\text{C}}{5.0 \times 10^{-4}\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}} = \frac{100}{5.0 \times 10^{-4}} = 200,000\text{ W}\cdot\text{m}^{-2} = 200\text{ kW}\cdot\text{m}^{-2}

.

At steady state, the heat flux across the composite structure is driven by the overall temperature gradient divided by total thermal resistance.

Anahtar Kavram

Thermal Resistance in Composite Layers (Conduction)

Tahmini Süre:2m 0s

Soru 37Soru →

An ideal gas occupies a volume of $0.05\text{ m}^3$ inside a rigid container. If the gas exerts a pressure of $2.4 \times 10^5\text{ N m}^{-2}$ on the walls of the container, what is the total translational kinetic energy of the gas molecules in joules?

Cevabı ve açıklamayı göster

Cevap: 18000

Cevap

The total translational kinetic energy of the gas molecules is

18000\text{ J}

.

According to the kinetic theory of gases, the pressure

P

of an ideal gas is related to its total translational kinetic energy

E_k

and volume

V

by

P = \frac{2}{3} \frac{E_k}{V}

. Rearranging for

E_k

gives

E_k = \frac{3}{2} P V

. Substituting

P = 2.4 \times 10^5\text{ N m}^{-2}

and

V = 0.05\text{ m}^3

yields

E_k = \frac{3}{2} \times 2.4 \times 10^5 \times 0.05 = 18000\text{ J}

.

Adım Adım Çözüm

1

Identify the relationship between gas pressure, volume, and translational kinetic energy from kinetic theory.

P = \frac{2}{3} \left(\frac{E_k}{V}\right) \implies E_k = \frac{3}{2} P V

From kinetic theory, pressure is two-thirds of the total translational kinetic energy per unit volume.

2

Substitute the provided numerical values into the equation.

E_k = \frac{3}{2} \times (2.4 \times 10^5\text{ N m}^{-2}) \times (0.05\text{ m}^3)

The given values are pressure

P = 2.4 \times 10^5\text{ N m}^{-2}

and volume

V = 0.05\text{ m}^3

.

3

Evaluate the expression to determine the numerical result.

E_k = 1.5 \times 12000 = 18000\text{ J}

Multiplying the values gives the energy in Joules.

Anahtar Kavram

Relationship between pressure, volume, and total translational kinetic energy of gas molecules (

E_k = \frac{3}{2} P V

).

Soru 38Soru →

A copper rod has an initial length of $100\text{ cm}$ at $25^\circ\text{C}$ . When its temperature is increased to $75^\circ\text{C}$ , its length becomes $100.085\text{ cm}$ . What is the area expansivity of copper in $10^{-5}\text{ K}^{-1}$ ?

Cevabı ve açıklamayı göster

Cevap: 3.4

Cevap

The area expansivity of copper is

3.4 \times 10^{-5}\text{ K}^{-1}

, giving a numerical value of

3.4

in units of

10^{-5}\text{ K}^{-1}

.

Linear expansivity is obtained as

\alpha = \frac{\Delta L}{L_0 \Delta T} = \frac{0.085\text{ cm}}{100\text{ cm} \times 50\text{ K}} = 1.7 \times 10^{-5}\text{ K}^{-1}

. Since area expansivity

\beta

is related to linear expansivity by

\beta = 2\alpha

, multiplying by

2

yields

3.4 \times 10^{-5}\text{ K}^{-1}

, or

3.4

in units of

10^{-5}\text{ K}^{-1}

.

Adım Adım Çözüm

1

Calculate the temperature change

\Delta T = 75^\circ\text{C} - 25^\circ\text{C} = 50\text{ K}

Thermal expansion depends directly on the change in temperature.

2

Find the change in length

\Delta L = 100.085\text{ cm} - 100\text{ cm} = 0.085\text{ cm}

The expansion is the difference between the final length and original length.

3

Determine linear expansivity (

\alpha

)

\alpha = \frac{\Delta L}{L_0 \Delta T} = \frac{0.085}{100 \times 50} = 1.7 \times 10^{-5}\text{ K}^{-1}

Linear expansivity defines fractional change in length per degree temperature change.

4

Compute area expansivity (

\beta

)

\beta = 2\alpha = 2 \times (1.7 \times 10^{-5}) = 3.4 \times 10^{-5}\text{ K}^{-1}

Area (superficial) expansivity is equal to twice the linear expansivity.

Anahtar Kavram

Relationship between linear expansivity (

\alpha

) and area expansivity (

\beta = 2\alpha

).

Soru 39Soru →

A glass flask is filled to a mark with a liquid at $0^\circ\text{C}$ . If the linear expansivity of the glass is $8.0 \times 10^{-6} \text{ K}^{-1}$ and the apparent cubic expansivity of the liquid is $1.80 \times 10^{-4} \text{ K}^{-1}$ , what is the real cubic expansivity of the liquid?

1.88 \times 10^{-4} \text{ K}^{-1}

2.04 \times 10^{-4} \text{ K}^{-1}

1.56 \times 10^{-4} \text{ K}^{-1}

1.72 \times 10^{-4} \text{ K}^{-1}

Cevabı ve açıklamayı göster

Cevap: $2.04 \times 10^{-4} \text{ K}^{-1}$

Cevap

The real cubic expansivity of the liquid is

2.04 \times 10^{-4} \text{ K}^{-1}

.

The real cubic expansivity of a liquid is given by

\gamma_r = \gamma_a + \gamma_v

. Since the container's linear expansivity

\alpha = 8.0 \times 10^{-6} \text{ K}^{-1}

, its cubic expansivity is

\gamma_v = 3\alpha = 2.40 \times 10^{-5} \text{ K}^{-1} = 0.24 \times 10^{-4} \text{ K}^{-1}

. Adding this to the apparent cubic expansivity

1.80 \times 10^{-4} \text{ K}^{-1}

gives

2.04 \times 10^{-4} \text{ K}^{-1}

.

Adım Adım Çözüm

1

Calculate the cubic expansivity of the glass vessel (

\gamma_v

).

\gamma_v = 3 \times \alpha = 3 \times (8.0 \times 10^{-6} \text{ K}^{-1}) = 2.40 \times 10^{-5} \text{ K}^{-1} = 0.24 \times 10^{-4} \text{ K}^{-1}

Cubic expansivity of a solid material is three times its linear expansivity.

2

Apply the relationship between real expansivity (

\gamma_r

), apparent expansivity (

\gamma_a

), and vessel expansivity (

\gamma_v

).

\gamma_r = \gamma_a + \gamma_v

The real expansion of a liquid is the sum of its observed (apparent) expansion and the expansion of the containing vessel.

3

Substitute the values into the formula to find

\gamma_r

.

\gamma_r = 1.80 \times 10^{-4} \text{ K}^{-1} + 0.24 \times 10^{-4} \text{ K}^{-1} = 2.04 \times 10^{-4} \text{ K}^{-1}

Adding the apparent cubic expansivity and the vessel's cubic expansivity yields the true (real) cubic expansivity.

Anahtar Kavram

Real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the cubic expansivity of the vessel (

\gamma_r = \gamma_a + 3\alpha_v

).

Soru 40Soru →

A copper container with an initial volume of $800 \text{ cm}^3$ at $25^\circ\text{C}$ is completely filled with oil. The real cubic expansivity of the oil is $6.5 \times 10^{-4} \text{ K}^{-1}$ and the linear expansivity of copper is $1.5 \times 10^{-5} \text{ K}^{-1}$ . What volume of oil (in $\text{cm}^3$ ) will overflow when the temperature of the system is raised to $75^\circ\text{C}$ ?

Cevabı ve açıklamayı göster

Cevap: 24.2

Cevap

The volume of oil that overflows is

24.2 \text{ cm}^3

.

When a container completely filled with liquid is heated, both liquid and container expand. The overflow volume equals the apparent volume expansion of the liquid

\Delta V_a = V_0 \gamma_a \Delta T

. The apparent cubic expansivity

\gamma_a

is obtained by subtracting the container's volume expansivity (

\gamma_v = 3\alpha = 4.5 \times 10^{-5} \text{ K}^{-1}

) from the liquid's real cubic expansivity (

\gamma_r = 6.5 \times 10^{-4} \text{ K}^{-1}

), giving

\gamma_a = 6.05 \times 10^{-4} \text{ K}^{-1}

. Multiplying by

V_0 = 800 \text{ cm}^3

and

\Delta T = 50 \text{ K}

gives

24.2 \text{ cm}^3

.

Adım Adım Çözüm

1

Calculate the cubic expansivity of the copper container

\gamma_v = 3 \alpha = 3 \times (1.5 \times 10^{-5} \text{ K}^{-1}) = 4.5 \times 10^{-5} \text{ K}^{-1} = 0.45 \times 10^{-4} \text{ K}^{-1}

The volumetric expansion coefficient of a solid container is three times its linear expansivity.

2

Determine the apparent cubic expansivity of the oil

\gamma_a = \gamma_r - \gamma_v = 6.5 \times 10^{-4} \text{ K}^{-1} - 0.45 \times 10^{-4} \text{ K}^{-1} = 6.05 \times 10^{-4} \text{ K}^{-1}

The apparent expansion of a liquid accounts for the concurrent thermal expansion of the containing vessel.

3

Compute the temperature increase

\Delta T = 75^\circ\text{C} - 25^\circ\text{C} = 50 \text{ K}

Temperature change is the final temperature minus the initial temperature.

4

Calculate the volume of oil that overflows

\Delta V_a = V_0 \gamma_a \Delta T = 800 \text{ cm}^3 \times (6.05 \times 10^{-4} \text{ K}^{-1}) \times 50 \text{ K} = 24.2 \text{ cm}^3

The overflow volume is equal to the apparent increase in volume of the liquid.

Anahtar Kavram

Apparent and Real Expansion of Liquids

Thermal Physics

Cevap

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