Thermal Physics Alıştırma Soruları — JAMB UTME

Match each thermometer type listed on the left with its corresponding thermometric property on the right.

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Cevap

Liquid-in-glass thermometer matches change in volume of a liquid column; Constant-volume gas thermometer matches change in gas pressure; Platinum resistance thermometer matches change in electrical resistance; Thermocouple matches change in electromotive force (e.m.f.).

Each thermometer relies on a physical property that changes linearly or predictably with temperature: liquid-in-glass uses volume expansion of liquid, constant-volume gas thermometer uses gas pressure variation, platinum resistance thermometer uses electrical resistance change, and thermocouple uses electromotive force generated across thermal junctions.

Adım Adım Çözüm

1

Identify the thermometric property for a liquid-in-glass thermometer.

Expansion of liquid volume.

The liquid (mercury or alcohol) expands up a narrow capillary tube as temperature increases.

2

Identify the thermometric property for a constant-volume gas thermometer.

Pressure of a gas.

At constant volume, the pressure of an ideal gas changes linearly with absolute temperature.

3

Identify the thermometric property for a platinum resistance thermometer.

Electrical resistance.

The electrical resistance of metals increases predictably with temperature.

4

Identify the thermometric property for a thermocouple.

Electromotive force (e.m.f.).

A temperature difference between two thermoelectric junctions induces a proportional voltage.

Anahtar Kavram

Thermometric properties of common thermometers

Tahmini Süre:45s

Soru 2Soru →

On a warm afternoon, the air temperature in a physics laboratory is $30^\circ\text{C}$ , where the saturated vapour pressure of water is $32.0\text{ mmHg}$ . When the air is cooled, condensation just begins to form on a metal vessel at $20^\circ\text{C}$ . Given that the saturated vapour pressure of water at $20^\circ\text{C}$ is $17.6\text{ mmHg}$ , what is the relative humidity of the air in percentage?

Cevabı ve açıklamayı göster

Cevap: 55

Cevap

The relative humidity of the air is

55\%

.

The dew point is the temperature at which condensation begins, indicating that the actual water vapour pressure present in the air equals the saturated vapour pressure at

20^\circ\text{C}

, which is

17.6\text{ mmHg}

. Dividing this actual vapour pressure by the saturated vapour pressure at the ambient air temperature of

30^\circ\text{C}

(

32.0\text{ mmHg}

) and multiplying by

100\%

yields

\frac{17.6}{32.0} \times 100\% = 55\%

.

Adım Adım Çözüm

1

Determine the actual vapour pressure in the air

Actual vapour pressure =

17.6\text{ mmHg}

Condensation starts at the dew point (

20^\circ\text{C}

), meaning the actual vapour pressure in the air equals the saturated vapour pressure at the dew point.

2

Determine the saturated vapour pressure at the air temperature

Saturated vapour pressure at

30^\circ\text{C}

=

32.0\text{ mmHg}

This is the maximum vapour pressure the air can exert at its current ambient temperature.

3

Compute the relative humidity percentage

\text{Relative Humidity} = \frac{17.6}{32.0} \times 100\% = 55\%

Relative humidity is defined as the ratio of actual vapour pressure to saturated vapour pressure at air temperature, expressed as a percentage.

Anahtar Kavram

Calculation of relative humidity from saturated vapour pressure at dew point and air temperature

Tahmini Süre:1m 30s

Soru 3Soru →

A solid brass cube with an edge length of $10\text{ cm}$ at $15^\circ\text{C}$ is heated to a temperature of $115^\circ\text{C}$ . If the linear expansivity of brass is $2.0 \times 10^{-5}\text{ K}^{-1}$ , what is the increase in the volume of the cube in $\text{cm}^3$ ?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The increase in the volume of the brass cube is

6.0\text{ cm}^3

.

The volume expansion of a solid is given by

\Delta V = V_1 \gamma \Delta T

. The initial volume of the cube is

V_1 = (10\text{ cm})^3 = 1000\text{ cm}^3

and the temperature change is

\Delta T = 115^\circ\text{C} - 15^\circ\text{C} = 100\text{ K}

. Because the expansion occurs in three dimensions, the volume expansivity is

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

. Substituting these values yields

\Delta V = 1000 \times 6.0 \times 10^{-5} \times 100 = 6.0\text{ cm}^3

.

Adım Adım Çözüm

1

Calculate the initial volume of the cube

V_1 = (10\text{ cm})^3 = 1000\text{ cm}^3

The volume of a cube is calculated using

V = L^3

, where

L

is the edge length.

2

Determine the change in temperature

\Delta T = 115^\circ\text{C} - 15^\circ\text{C} = 100\text{ K}

The change in temperature is the difference between the final and initial temperatures.

3

Calculate the volume expansivity (cubic expansivity)

\gamma = 3\alpha = 3 \times (2.0 \times 10^{-5}\text{ K}^{-1}) = 6.0 \times 10^{-5}\text{ K}^{-1}

Volume expansivity

\gamma

is three times the linear expansivity

\alpha

for isotropic solids.

4

Compute the increase in volume

\Delta V = V_1 \gamma \Delta T = 1000 \times (6.0 \times 10^{-5}) \times 100 = 6.0\text{ cm}^3

The formula for volume expansion is

\Delta V = V_1 \gamma \Delta T

.

Anahtar Kavram

Thermal expansion of solids: volume expansion (

\Delta V = V_1 \gamma \Delta T

) and the relationship between linear and volume expansivity (

\gamma = 3\alpha

).

Soru 4Soru →

An iron ring has an internal cross-sectional area of $0.50\text{ m}^2$ at $30^\circ\text{C}$ . If the linear expansivity of iron is $1.2 \times 10^{-5}\text{ K}^{-1}$ , what is the increase in its internal cross-sectional area when heated to $130^\circ\text{C}$ ?

1.2 \times 10^{-3}\text{ m}^2

6.0 \times 10^{-4}\text{ m}^2

1.8 \times 10^{-3}\text{ m}^2

2.4 \times 10^{-3}\text{ m}^2

Cevabı ve açıklamayı göster

Cevap: $1.2 \times 10^{-3}\text{ m}^2$

Cevap

The increase in the internal cross-sectional area of the ring is

1.2 \times 10^{-3}\text{ m}^2

.

For two-dimensional (area) expansion of solids, the area expansivity

\beta

is equal to twice the linear expansivity (

2\alpha

). Given

\alpha = 1.2 \times 10^{-5}\text{ K}^{-1}

,

\beta = 2.4 \times 10^{-5}\text{ K}^{-1}

. Multiplying by the initial area (

0.50\text{ m}^2

) and temperature rise (

100\text{ K}

) yields an area increase of

1.2 \times 10^{-3}\text{ m}^2

.

Adım Adım Çözüm

1

Determine the temperature change (

\Delta T

) and the superficial expansivity (

\beta

).

\Delta T = 130^\circ\text{C} - 30^\circ\text{C} = 100\text{ K}

, and

\beta = 2\alpha = 2 \times (1.2 \times 10^{-5}\text{ K}^{-1}) = 2.4 \times 10^{-5}\text{ K}^{-1}

.

Area expansion depends on superficial expansivity, which is twice the linear expansivity for an isotropic solid.

2

Calculate the increase in area (

\Delta A

) using the area expansion formula.

\Delta A = A_0 \beta \Delta T = 0.50\text{ m}^2 \times (2.4 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 1.2 \times 10^{-3}\text{ m}^2

.

The fractional change in area is directly proportional to initial area, superficial expansivity, and temperature change.

Anahtar Kavram

Relationship between linear expansivity (

\alpha

) and superficial expansivity (

\beta = 2\alpha

) in thermal expansion of area.

Tahmini Süre:1m 30s

Soru 5Soru →

An aerosol canister containing an ideal gas at an initial pressure of $2.20 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ is accidentally thrown into a fire, causing the temperature of the gas to rise to $327^\circ\text{C}$ . Assuming the volume of the canister remains constant, what is the new pressure of the gas inside the canister?

1.10 \times 10^5\text{ Pa}

2.66 \times 10^6\text{ Pa}

4.40 \times 10^5\text{ Pa}

2.45 \times 10^5\text{ Pa}

Cevabı ve açıklamayı göster

Cevap: $4.40 \times 10^5\text{ Pa}$

Cevap

The new pressure of the gas inside the canister is

4.40 \times 10^5\text{ Pa}

.

According to Pressure's Law, the pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature in Kelvin (

\frac{P_1}{T_1} = \frac{P_2}{T_2}

). Converting the temperatures gives

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

and

T_2 = 327 + 273 = 600\text{ K}

. Multiplying the initial pressure by the ratio

\frac{600\text{ K}}{300\text{ K}} = 2

gives

4.40 \times 10^5\text{ Pa}

.

Adım Adım Çözüm

1

Convert initial and final temperatures from degrees Celsius to Kelvin.

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

and

T_2 = 327^\circ\text{C} + 273 = 600\text{ K}

.

All calculations using gas laws require absolute temperature measured in Kelvin.

2

State and rearrange Pressure's Law (Gay-Lussac's Law) for constant volume.

\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \times \frac{T_2}{T_1}

.

For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature.

3

Substitute the known values to calculate the final pressure

P_2

.

P_2 = 2.20 \times 10^5\text{ Pa} \times \frac{600\text{ K}}{300\text{ K}} = 4.40 \times 10^5\text{ Pa}

.

Since absolute temperature doubles from

300\text{ K}

to

600\text{ K}

, the final pressure must also double.

Anahtar Kavram

Pressure Law (Gay-Lussac's Law) and Absolute Temperature Conversion

Tahmini Süre:1m 30s

Soru 6Soru →

A diver releases a bubble of air of volume $8.00\text{ cm}^3$ at a depth where the water pressure is $3.50 \times 10^5\text{ Pa}$ and the temperature is $7^\circ\text{C}$ . What is the volume of the air bubble, in $\text{cm}^3$ , just as it reaches the surface where the pressure is $1.00 \times 10^5\text{ Pa}$ and the temperature is $27^\circ\text{C}$ ?

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

The final volume of the air bubble at the surface is

30.0\text{ cm}^3

.

Using the combined gas law

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

with absolute temperatures

T_1 = 280\text{ K}

and

T_2 = 300\text{ K}

yields a final volume of

30.0\text{ cm}^3

.

Adım Adım Çözüm

1

Convert temperatures from Celsius to Kelvin

T_1 = 7^\circ\text{C} + 273 = 280\text{ K}

and

T_2 = 27^\circ\text{C} + 273 = 300\text{ K}

Gas laws require absolute thermodynamic temperature in Kelvin.

2

Apply the combined gas law equation

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

The amount of gas in the bubble remains constant while pressure, volume, and temperature all change simultaneously.

3

Rearrange for the unknown volume

V_2

and substitute the values

V_2 = \frac{3.50 \times 10^5 \times 8.00 \times 300}{1.00 \times 10^5 \times 280} = 30.0\text{ cm}^3

Calculates the expanded volume of the air bubble at surface conditions.

Anahtar Kavram

Combined Gas Law

Soru 7Soru →

A gas sample enclosed in a container has an initial root-mean-square (r.m.s.) speed of $400\text{ m s}^{-1}$ at a temperature of $27^\circ\text{C}$ . If the gas is heated at constant volume until its temperature reaches $927^\circ\text{C}$ , what is the new r.m.s. speed of the gas molecules?

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

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

1600\text{ m s}^{-1}

2400\text{ m s}^{-1}

Cevabı ve açıklamayı göster

Cevap: $800\text{ m s}^{-1}$

Cevap

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

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

.

According to the kinetic theory of gases, the root-mean-square speed is directly proportional to the square root of the absolute temperature (

v_{\text{rms}} = \sqrt{3RT/M}

). Converting the temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 1200\text{ K}

. The ratio of absolute temperatures is

1200 / 300 = 4

. Taking the square root gives a factor of

2

, so the new r.m.s. speed is

400\text{ m s}^{-1} \times 2 = 800\text{ m s}^{-1}

.

Adım Adım Çözüm

1

Convert initial and final temperatures from Celsius to Kelvin

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

and

T_2 = 927 + 273 = 1200\text{ K}

Kinetic theory equations require absolute temperature in Kelvin.

2

Apply the relationship between r.m.s. speed and absolute temperature

v_{\text{rms}} \propto \sqrt{T} \implies \frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}

The mean kinetic energy of gas molecules is directly proportional to absolute temperature.

3

Substitute the values and calculate the final speed

v_2

v_2 = 400 \times \sqrt{\frac{1200}{300}} = 400 \times \sqrt{4} = 400 \times 2 = 800\text{ m s}^{-1}

Evaluating the square root factor yields the updated r.m.s. speed.

Anahtar Kavram

Root-mean-square speed of gas molecules is directly proportional to the square root of absolute temperature (

v_{\text{rms}} \propto \sqrt{T}

).

Tahmini Süre:1m 0s

Soru 8Soru →

Match each vacuum flask component or surface feature on the left with its primary mechanism for controlling heat transfer on the right.

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Cevap

Silvered inner surfaces match with radiation reflection; evacuated space matches with elimination of conduction and convection; cork stopper matches with prevention of convection and conduction at the opening; dull black casing matches with maximizing thermal radiation emission/absorption.

Each feature of the vacuum flask targets a specific heat transfer mode: silvering reflects infrared radiation; the vacuum eliminates particle-dependent transfer (conduction and convection); cork acts as an insulator preventing convection and conduction at the top; and black surfaces maximize thermal radiation emission/absorption.

Adım Adım Çözüm

1

Analyze the silvered inner walls of a vacuum flask

Silver surfaces are good reflectors of heat rays (infrared waves).

Radiant heat travels via electromagnetic waves and is reflected by shiny metallic coatings, minimizing radiation heat loss.

2

Analyze the vacuum space between the glass walls

Conduction and convection cannot occur across a vacuum.

Both conduction (particle vibration/electron flow) and convection (fluid movement) strictly require a physical medium.

3

Analyze the cork/plastic stopper

Cork prevents hot air circulation and thermal conduction across the opening.

Cork is a poor conductor of heat and stops evaporative/convective air currents from leaving the container.

4

Analyze the dull black exterior

Dull black surfaces are efficient radiation emitters/absorbers.

According to radiation principles, black matte surfaces radiate energy much faster than polished surfaces.

Anahtar Kavram

Modes of Heat Transfer and Practical Applications in Thermal Insulation

Soru 9Soru →

A resistance thermometer has a resistance of $4.0\,\Omega$ at the ice point ( $0^\circ\text{C}$ ) and $6.0\,\Omega$ at the steam point ( $100^\circ\text{C}$ ). When immersed in a liquid bath, its resistance is measured to be $5.2\,\Omega$ . What is the temperature of the bath?

60.0^\circ\text{C}

260.0^\circ\text{C}

20.0^\circ\text{C}

40.0^\circ\text{C}

Cevabı ve açıklamayı göster

Cevap: $60.0^\circ\text{C}$

Cevap

60.0^\circ\text{C}

The temperature of the bath is found by calculating the fraction of resistance change relative to the total change between the ice point and steam point:

\theta = \frac{5.2 - 4.0}{6.0 - 4.0} \times 100^\circ\text{C} = 60.0^\circ\text{C}

.

Adım Adım Çözüm

1

Identify the given thermometric parameters

R_0 = 4.0\,\Omega

,

R_{100} = 6.0\,\Omega

, and

R_\theta = 5.2\,\Omega

These are the measured resistance values corresponding to the lower fixed point, upper fixed point, and unknown temperature.

2

Apply the linear temperature interpolation formula for a resistance thermometer

\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}

Temperature on the Celsius scale is proportional to the relative change in the thermometric property between fixed points.

3

Substitute the values and compute the temperature

\theta = \frac{5.2 - 4.0}{6.0 - 4.0} \times 100^\circ\text{C} = \frac{1.2}{2.0} \times 100^\circ\text{C} = 60.0^\circ\text{C}

Evaluating the expression yields the exact temperature of the liquid bath.

Anahtar Kavram

Linear interpolation on temperature scales using thermometric properties

Soru 10Soru →

The length of the mercury column in an uncalibrated liquid-in-glass thermometer is $2.0\,\text{cm}$ at the ice point ( $0^\circ\text{C}$ ) and $18.0\,\text{cm}$ at the steam point ( $100^\circ\text{C}$ ). What is the temperature when the length of the mercury column is $9.2\,\text{cm}$ ?

40^\circ\text{C}

45^\circ\text{C}

55^\circ\text{C}

57.5^\circ\text{C}

Cevabı ve açıklamayı göster

Cevap: $45^\circ\text{C}$

Cevap

The temperature corresponding to a mercury column length of

9.2\,\text{cm}

is

45^\circ\text{C}

.

The temperature on the Celsius scale is given by the formula

\theta = \frac{L_\theta - L_0}{L_{100} - L_0} \times 100^\circ\text{C}

. Substituting

L_0 = 2.0\,\text{cm}

,

L_{100} = 18.0\,\text{cm}

, and

L_\theta = 9.2\,\text{cm}

yields

\theta = \frac{7.2}{16.0} \times 100^\circ\text{C} = 45^\circ\text{C}

.

Adım Adım Çözüm

1

Identify the fixed points and the thermometric property values.

Ice point length

L_0 = 2.0\,\text{cm}

, steam point length

L_{100} = 18.0\,\text{cm}

, and observed length

L_\theta = 9.2\,\text{cm}

.

The linear scale equation requires establishing the reference points on the Celsius scale.

2

Calculate the fundamental interval length (

L_{100} - L_0

).

L_{100} - L_0 = 18.0\,\text{cm} - 2.0\,\text{cm} = 16.0\,\text{cm}

.

The fundamental interval represents the total change in length corresponding to

100^\circ\text{C}

.

3

Apply the linear interpolation formula

\theta = \frac{L_\theta - L_0}{L_{100} - L_0} \times 100^\circ\text{C}

.

\theta = \frac{9.2 - 2.0}{16.0} \times 100^\circ\text{C} = \frac{7.2}{16.0} \times 100^\circ\text{C} = 45^\circ\text{C}

.

This scales the fractional change in thermometric property above the ice point to degrees Celsius.

Anahtar Kavram

Linear interpolation on empirical temperature scales

Tahmini Süre:1m 30s

Soru 11Soru →

A liquid will boil when its saturated vapour pressure becomes equal to the prevailing external atmospheric pressure.

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

The statement is True. A liquid boils when its saturated vapour pressure equals the external atmospheric pressure.

The statement accurately expresses the fundamental thermodynamic condition for boiling: the temperature of the liquid must reach a point where its saturated vapour pressure equals the surrounding atmospheric pressure.

Adım Adım Çözüm

1

Recall the definition of boiling point in thermal physics.

Boiling is the rapid conversion of liquid into gas occurring throughout the liquid body.

To determine the exact physical condition required for boiling to take place.

2

Relate saturated vapour pressure (SVP) to atmospheric pressure.

Bubbles of vapour can form within the liquid only when the pressure inside the bubbles (SVP) is equal to or greater than the pressure pushing down from the outside atmosphere.

If SVP is lower than atmospheric pressure, any vapour bubble attempting to form inside the liquid will immediately collapse.

Anahtar Kavram

Condition for Boiling of Liquids

Soru 12Soru →

What quantity of heat energy is required to completely melt $0.50\text{ kg}$ of ice at $0^\circ\text{C}$ into water at $0^\circ\text{C}$ ? (Specific latent heat of fusion of ice = $3.3 \times 10^5\text{ J kg}^{-1}$ )

1.65 \times 10^5\text{ J}

3.30 \times 10^5\text{ J}

6.60 \times 10^5\text{ J}

1.65 \times 10^7\text{ J}

Cevabı ve açıklamayı göster

Cevap: $1.65 \times 10^5\text{ J}$

Cevap

The quantity of heat energy required is

1.65 \times 10^5\text{ J}

.

The quantity of heat required for melting at constant temperature depends only on mass and specific latent heat of fusion:

Q = mL_f = 0.50 \times 3.3 \times 10^5 = 1.65 \times 10^5\text{ J}

.

Adım Adım Çözüm

1

Identify the given values and formula for change of state

Mass

m = 0.50\text{ kg}

, Specific latent heat of fusion

L_f = 3.3 \times 10^5\text{ J kg}^{-1}

. The formula is

Q = m L_f

.

During a phase change at constant temperature, latent heat is absorbed without a change in temperature.

2

Calculate total heat energy

Q

Q = 0.50\text{ kg} \times 3.3 \times 10^5\text{ J kg}^{-1} = 1.65 \times 10^5\text{ J}

.

Multiplying mass by the specific latent heat gives the total energy transferred.

Anahtar Kavram

Latent Heat of Fusion

Tahmini Süre:45s

Soru 13Soru →

A copper calorimeter of heat capacity $300\text{ J K}^{-1}$ contains $0.5\text{ kg}$ of water at an initial temperature of $25^\circ\text{C}$ . An electric heater rated at $800\text{ W}$ is immersed in the water to heat the system for $4\text{ minutes}$ . If heat is lost to the surrounding environment at a constant rate of $200\text{ W}$ throughout the heating duration, what is the final temperature of the water-calorimeter system in $^\circ\text{C}$ ? (Take the specific heat capacity of water as $4200\text{ J kg}^{-1}\text{K}^{-1}$ )

Cevabı ve açıklamayı göster

Cevap: 85

Cevap

The final temperature of the system is

85^\circ\text{C}

.

The net thermal energy added to the system accounts for both the supplied electrical energy and the heat energy lost to the surroundings:

Q_{\text{net}} = (800 - 200)\text{ W} \times 240\text{ s} = 144,000\text{ J}

. The total heat capacity of the water-calorimeter system is

C_{\text{total}} = 300\text{ J K}^{-1} + (0.5\text{ kg} \times 4200\text{ J kg}^{-1}\text{K}^{-1}) = 2400\text{ J K}^{-1}

. The temperature increase is

\Delta T = \frac{144,000}{2400} = 60^\circ\text{C}

. Adding this to the initial temperature of

25^\circ\text{C}

gives the final temperature of

85^\circ\text{C}

.

Adım Adım Çözüm

1

Convert heating time to standard SI units (seconds)

t = 4 \times 60\text{ s} = 240\text{ s}

Power is measured in Joules per second (Watts), so time must be in seconds.

2

Calculate the net rate of heat energy input to the system

P_{\text{net}} = 800\text{ W} - 200\text{ W} = 600\text{ W}

The net heating power is the input power minus the power dissipated as heat loss.

3

Calculate total net heat energy transferred to the system

Q_{\text{net}} = P_{\text{net}} \times t = 600\text{ W} \times 240\text{ s} = 144,000\text{ J}

Thermal energy transferred equals net power multiplied by time.

4

Compute the total heat capacity of the combined system (water + calorimeter)

C_{\text{total}} = C_{\text{calorimeter}} + (m_{\text{water}} \times c_{\text{water}}) = 300\text{ J K}^{-1} + (0.5\text{ kg} \times 4200\text{ J kg}^{-1}\text{K}^{-1}) = 2400\text{ J K}^{-1}

Heat capacity of water is mass multiplied by specific heat capacity, added to the calorimeter's given heat capacity.

5

Calculate the temperature rise of the system

\Delta T = \frac{Q_{\text{net}}}{C_{\text{total}}} = \frac{144,000\text{ J}}{2400\text{ J K}^{-1}} = 60^\circ\text{C}

Temperature change is total net heat supplied divided by total heat capacity.

6

Find the final temperature of the system

T_{\text{final}} = T_{\text{initial}} + \Delta T = 25^\circ\text{C} + 60^\circ\text{C} = 85^\circ\text{C}

Final temperature equals initial temperature plus temperature increase.

Anahtar Kavram

Conservation of thermal energy in calorimeter systems with continuous power loss

Tahmini Süre:1m 30s

Soru 14Soru →

A gas cylinder fitted with a frictionless piston contains a fixed mass of ideal gas occupying a volume of $0.040\text{ m}^3$ at a pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . The gas is heated to $127^\circ\text{C}$ while expanding to a new volume of $0.080\text{ m}^3$ . What is the final pressure of the gas?

1.00 \times 10^5\text{ Pa}

3.53 \times 10^5\text{ Pa}

7.50 \times 10^4\text{ Pa}

4.00 \times 10^5\text{ Pa}

Cevabı ve açıklamayı göster

Cevap: $1.00 \times 10^5\text{ Pa}$

Cevap

The final pressure of the gas is

1.00 \times 10^5\text{ Pa}

.

By converting temperatures to absolute zero scale (

300\text{ K}

and

400\text{ K}

) and applying the Combined Gas Law

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

, the final pressure

P_2

evaluates directly to

1.00 \times 10^5\text{ Pa}

.

Adım Adım Çözüm

1

Convert temperatures from Celsius to Kelvin.

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

and

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

.

Gas laws require absolute temperatures measured on the Kelvin scale.

2

Apply the Combined Gas Law formula

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

.

\frac{(1.50 \times 10^5\text{ Pa})(0.040\text{ m}^3)}{300\text{ K}} = \frac{P_2 (0.080\text{ m}^3)}{400\text{ K}}

.

The mass of the gas is fixed while pressure, volume, and temperature all change.

3

Rearrange and solve for final pressure

P_2

.

P_2 = 1.50 \times 10^5 \times \left(\frac{0.040}{0.080}\right) \times \left(\frac{400}{300}\right) = 1.00 \times 10^5\text{ Pa}

.

Simplifying the numerical expression yields the final equilibrium pressure.

Anahtar Kavram

Combined Gas Law

Tahmini Süre:2m 0s

Soru 15Soru →

An electric heater rated at $50\text{ W}$ is used to heat a solid block of mass $1.5\text{ kg}$ for $4\text{ minutes}$ . During this period, the temperature of the block increases from $30^\circ\text{C}$ to $70^\circ\text{C}$ . If $20\%$ of the heat energy supplied by the heater is lost to the surroundings, what is the heat capacity of the block?

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $240\text{ J K}^{-1}$

Cevap

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

The correct answer is

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

. The heater delivers

12{}000\text{ J}

of energy in

4\text{ minutes}

. Accounting for

20\%

environmental heat loss leaves

9{}600\text{ J}

absorbed by the block. Dividing this energy by the

40\text{ K}

temperature rise gives a total heat capacity of

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

.

Adım Adım Çözüm

1

Calculate total electrical energy supplied by heater

Q_{\text{supplied}} = P \times t = 50\text{ W} \times (4 \times 60\text{ s}) = 12{}000\text{ J}

Electrical work converted to heat is given by power multiplied by time in seconds.

2

Determine useful heat absorbed by block after accounting for energy loss

Q_{\text{useful}} = (1 - 0.20) \times 12{}000\text{ J} = 0.80 \times 12{}000\text{ J} = 9{}600\text{ J}

Since

20\%

of supplied energy is lost,

80\%

is retained to raise the temperature of the block.

3

Calculate temperature rise

\Delta T = 70^\circ\text{C} - 30^\circ\text{C} = 40\text{ K}

Temperature difference is the final temperature minus the initial temperature.

4

Calculate heat capacity of the block

C = \frac{Q_{\text{useful}}}{\Delta T} = \frac{9{}600\text{ J}}{40\text{ K}} = 240\text{ J K}^{-1}

Heat capacity

C

is defined as total heat absorbed per unit temperature change (

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

).

Anahtar Kavram

Heat Capacity (

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

) represents total thermal capacity of a body, whereas Specific Heat Capacity (

c = \frac{Q}{m\Delta T}

) is heat capacity per unit mass.

Soru 16Soru →

A metal rod has a linear expansivity of $1.5 \times 10^{-5}\text{ K}^{-1}$ . What is the volume expansivity of a sphere made from the same metal?

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $4.5 \times 10^{-5}\text{ K}^{-1}$

Cevap

The volume expansivity of the sphere is

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

.

The volume expansivity

\gamma

of a uniform solid object is related to its linear expansivity

\alpha

by

\gamma = 3\alpha

. Multiplying

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

by

3

gives

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

.

Adım Adım Çözüm

1

Identify the mathematical relationship between linear expansivity (

\alpha

) and volume (cubical) expansivity (

\gamma

).

\gamma = 3\alpha

For an isotropic solid, volume expansion occurs equally in three dimensions, making the volume coefficient three times the linear coefficient.

2

Substitute the given value of linear expansivity into the formula and calculate.

\gamma = 3 \times (1.5 \times 10^{-5}\text{ K}^{-1}) = 4.5 \times 10^{-5}\text{ K}^{-1}

Obtain the numeric value of volume expansivity.

Anahtar Kavram

Relationship between linear and volume expansivity of solids

Soru 17Soru →

A glass vessel has a linear expansivity of $1.0 \times 10^{-5} \text{ K}^{-1}$ and is filled with a liquid. If the apparent cubic expansivity of the liquid in this vessel is $1.5 \times 10^{-4} \text{ K}^{-1}$ , calculate the real cubic expansivity of the liquid in units of $10^{-4} \text{ K}^{-1}$ .

Cevabı ve açıklamayı göster

Cevap: 1.8

Cevap

The real cubic expansivity of the liquid is

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

(giving

1.8

in units of

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

).

The real cubic expansivity of a liquid equals the sum of its apparent cubic expansivity and the volume expansivity of the vessel (\gamma_r = \gamma_a + \gamma_v). First, convert the linear expansivity of the glass vessel to volume expansivity:

\gamma_v = 3\alpha = 3 \times 1.0 \times 10^{-5} \text{ K}^{-1} = 0.3 \times 10^{-4} \text{ K}^{-1}

. Adding this to the apparent cubic expansivity (

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

) yields a real cubic expansivity of

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

.

Adım Adım Çözüm

1

Determine the cubic expansivity of the vessel (\gamma_v)

\gamma_v = 3.0 \times 10^{-5} \text{ K}^{-1} = 0.3 \times 10^{-4} \text{ K}^{-1}

The volume (cubic) expansivity of a solid vessel is three times its linear expansivity (\gamma_v = 3\alpha).

2

Calculate the real cubic expansivity of the liquid (\gamma_r)

\gamma_r = 1.8 \times 10^{-4} \text{ K}^{-1}

Real cubic expansivity is the sum of apparent cubic expansivity and vessel cubic expansivity (\gamma_r = \gamma_a + \gamma_v).

Anahtar Kavram

Real and Apparent Cubic Expansivity of Liquids

Soru 18Soru →

A rigid steel container holds a fixed mass of gas at an initial pressure of $1.20 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . If the temperature of the gas is increased to $127^\circ\text{C}$ while keeping its volume constant, what is the final pressure of the gas?

1.60 \times 10^5\text{ Pa}

5.64 \times 10^5\text{ Pa}

0.90 \times 10^5\text{ Pa}

4.44 \times 10^5\text{ Pa}

Cevabı ve açıklamayı göster

Cevap: $1.60 \times 10^5\text{ Pa}$

Cevap

The final pressure of the gas is

1.60 \times 10^5\text{ Pa}

.

According to Gay-Lussac's law, at constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature (

P_1/T_1 = P_2/T_2

). Converting the temperatures to Kelvin gives

T_1 = 300\text{ K}

and

T_2 = 400\text{ K}

. Solving for

P_2

yields

P_2 = 1.20 \times 10^5\text{ Pa} \times \frac{400}{300} = 1.60 \times 10^5\text{ Pa}

.

Adım Adım Çözüm

1

Convert temperatures from Celsius to Kelvin

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

,

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

Gas laws require absolute temperature in Kelvin to maintain proportional relationships.

2

Apply Pressure Law (Gay-Lussac's Law) for constant volume

\frac{P_1}{T_1} = \frac{P_2}{T_2} \implies P_2 = P_1 \times \frac{T_2}{T_1}

For a fixed mass of gas at constant volume, pressure is directly proportional to absolute temperature.

3

Substitute the known values to calculate the final pressure

P_2 = (1.20 \times 10^5\text{ Pa}) \times \frac{400\text{ K}}{300\text{ K}} = 1.60 \times 10^5\text{ Pa}

Multiplying the initial pressure by the temperature expansion factor yields the final pressure.

Anahtar Kavram

Gay-Lussac's Law (Pressure Law)

Soru 19Soru →

An electric heater rated at $1.0\text{ kW}$ is used to convert $0.50\text{ kg}$ of ice initially at $-10^\circ\text{C}$ completely into water at $50^\circ\text{C}$ . Assuming zero thermal energy loss to the surroundings, what is the total time required for this conversion?

(Take specific heat capacity of ice = $2100\text{ J kg}^{-1}\text{ K}^{-1}$ , specific latent heat of fusion of ice = $3.36 \times 10^5\text{ J kg}^{-1}$ , specific heat capacity of water = $4200\text{ J kg}^{-1}\text{ K}^{-1}$ )

283.5\text{ s}

273.0\text{ s}

1953.5\text{ s}

126.0\text{ s}

Cevabı ve açıklamayı göster

Cevap: $283.5\text{ s}$

Cevap

283.5\text{ s}

The correct response of

283.5\text{ s}

accurately accounts for all three distinct phases of thermal absorption: raising the temperature of solid ice from

-10^\circ\text{C}

to

0^\circ\text{C}

(

10,500\text{ J}

), melting the ice to water at constant temperature (

168,000\text{ J}

), and raising the liquid water temperature to

50^\circ\text{C}

(

105,000\text{ J}

). Dividing the total energy of

283,500\text{ J}

by the heater power of

1000\text{ W}

yields

283.5\text{ s}

.

Adım Adım Çözüm

1

Calculate the heat required to raise the temperature of ice from

-10^\circ\text{C}

to

0^\circ\text{C}

(

Q_1

)

Q_1 = m \cdot c_{\text{ice}} \cdot \Delta T_1 = 0.50 \times 2100 \times (0 - (-10)) = 10,500\text{ J}

Ice must reach its melting point at

0^\circ\text{C}

before any phase change can occur.

2

Calculate the latent heat required to melt ice at

0^\circ\text{C}

into water at

0^\circ\text{C}

(

Q_2

)

Q_2 = m \cdot L_f = 0.50 \times 3.36 \times 10^5 = 168,000\text{ J}

Phase change occurs at a constant temperature of

0^\circ\text{C}

using latent heat of fusion.

3

Calculate the heat required to raise the temperature of the resulting water from

0^\circ\text{C}

to

50^\circ\text{C}

(

Q_3

)

Q_3 = m \cdot c_{\text{water}} \cdot \Delta T_2 = 0.50 \times 4200 \times (50 - 0) = 105,000\text{ J}

Once completely melted, sensible heat is absorbed by liquid water up to the target temperature.

4

Calculate total heat energy required (

Q_{\text{total}}

) and convert power to watts

Q_{\text{total}} = 10,500 + 168,000 + 105,000 = 283,500\text{ J}

, and

P = 1.0\text{ kW} = 1000\text{ W}

Total energy is the sum of all individual stage energies.

5

Determine the time required (

t

)

t = \frac{Q_{\text{total}}}{P} = \frac{283,500}{1000} = 283.5\text{ s}

Power is defined as energy per unit time (

P = \frac{Q}{t}

).

Anahtar Kavram

Multi-stage thermal energy balance combining sensible heat (

Q = m c \Delta T

) and latent heat of fusion (

Q = m L_f

).

Alternatif Yöntem

Calculate energy per unit mass first:

q_{\text{total}} = c_{\text{ice}}\Delta T_1 + L_f + c_{\text{water}}\Delta T_2 = (2100 \times 10) + 336000 + (4200 \times 50) = 567,000\text{ J kg}^{-1}

. Then total energy

Q = 0.50 \times 567,000 = 283,500\text{ J}

, leading to

t = \frac{283,500}{1000} = 283.5\text{ s}

.

Tahmini Süre:3m 0s

Soru 20Soru →

Match each experimental temperature measurement requirement on the left with the most appropriate thermometric instrument on the right based on its thermometric property and operational characteristics.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

The correct pairings match each measurement requirement to its corresponding thermometric instrument based on its fundamental thermometric property: standard reference calibration pairs with the constant-volume gas thermometer; high-precision steady measurement pairs with the platinum resistance thermometer; rapid localized temperature change measurement pairs with the thermocouple; and non-contact high-temperature measurement pairs with the optical pyrometer.

Each instrument is correctly matched according to the specific physical property that changes measurably with temperature (

P

,

R

, e.m.f., and radiation intensity) and its operational suitability.

Adım Adım Çözüm

1

Analyze requirement 1: standard reference calibration using gas pressure at constant volume.

Identified thermometric property as pressure

P

at constant volume

V

, which defines the constant-volume gas thermometer.

Gas thermometers closely approximate the absolute thermodynamic scale and serve as calibration standards.

2

Analyze requirement 2: high-precision steady measurement using resistance variation with thermal lag.

Identified thermometric property as electrical resistance

R

, which corresponds to the platinum resistance thermometer.

Platinum wire resistance changes predictably with temperature, providing high accuracy for stable temperatures.

3

Analyze requirement 3: rapid localized temperature measurement via thermal e.m.f.

Identified thermometric property as thermoelectric voltage (e.m.f.), which corresponds to the thermocouple.

The small thermal mass of thermocouple junctions allows low response times for fast transient measurements.

4

Analyze requirement 4: non-contact measurement of glowing bodies using radiation.

Identified physical principle as thermal radiation intensity, corresponding to the optical pyrometer.

Pyrometers detect infrared/visible radiation, avoiding structural melting associated with direct contact at extreme temperatures.

Anahtar Kavram

Thermometric Properties and Operational Limits of Thermometers

Tahmini Süre:2m 0s

Thermal Physics

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