All practice questions

Question 1421Question →

Calculate the thermal energy, in joules, required to completely melt $0.50\text{ kg}$ of ice at its melting point of $0^\circ\text{C}$ . (Take the specific latent heat of fusion of ice as $3.36 \times 10^5\text{ J kg}^{-1}$ ).

Show answer & explanation

Answer: 168000

Answer

The total heat required to completely melt the ice is

168,000\text{ J}

.

During a state change from solid to liquid at constant temperature, the thermal energy absorbed is given directly by

Q = m L_f

. Multiplying mass (

0.50\text{ kg}

) by specific latent heat (

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

) gives

168,000\text{ J}

.

Step-by-Step Solution

1

Identify the given values and formula

Mass

m = 0.50\text{ kg}

, Specific latent heat of fusion

L_f = 3.36 \times 10^5\text{ J kg}^{-1}

, Formula:

Q = m L_f

Since the phase change occurs at constant temperature (

0^\circ\text{C}

), sensible heat calculation (

mc\Delta T

) is zero and only latent heat is needed.

2

Substitute values into the formula and calculate

Q = 0.50\text{ kg} \times 3.36 \times 10^5\text{ J kg}^{-1} = 168,000\text{ J}

Direct multiplication of mass and specific latent heat yields the energy required in Joules.

Key Concept

Specific Latent Heat of Fusion

Question 1422Question →

A series alternating current (AC) circuit consists of a resistor of resistance $R = 40\ \Omega$ , an inductor with inductive reactance $X_L = 70\ \Omega$ , and a capacitor with capacitive reactance $X_C = 40\ \Omega$ , connected to an AC voltage supply of $100\ \text{V}$ . What is the power factor of the circuit?

Show answer & explanation

Answer: 0.8

Answer

The power factor of the circuit is 0.8.

The total impedance

Z

of the series circuit is found using

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (70 - 40)^2} = 50\ \Omega

. The power factor is the ratio of resistance to impedance:

\cos\phi = \frac{R}{Z} = \frac{40}{50} = 0.8

.

Step-by-Step Solution

1

Calculate the net reactance of the circuit.

X = X_L - X_C = 70\ \Omega - 40\ \Omega = 30\ \Omega

In a series RLC circuit, the net reactance is the arithmetic difference between the inductive reactance and the capacitive reactance.

2

Calculate the total impedance of the circuit.

Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + 30^2} = \sqrt{2500} = 50\ \Omega

Impedance represents the combined opposition to current flow from resistance and net reactance in quadrature.

3

Determine the power factor of the circuit.

\cos\phi = \frac{R}{Z} = \frac{40}{50} = 0.8

The power factor is equal to the cosine of the phase angle, which is defined as the ratio of resistance to total impedance.

Key Concept

Power Factor in AC Circuits

Estimated Time:1m 30s

Question 1423Question →

A body of mass $0.50\text{ kg}$ connected to a light helical spring of force constant $32\text{ N/m}$ executes simple harmonic motion on a smooth horizontal surface. If the total mechanical energy of the oscillating system is $0.16\text{ J}$ , what is the maximum speed of the body in $\text{m/s}$ ?

Show answer & explanation

Answer: 0.8

Answer

The maximum speed of the body is

0.8\text{ m/s}

.

The total mechanical energy in simple harmonic motion is equal to the maximum kinetic energy at the equilibrium position:

E = \frac{1}{2} m v_{\text{max}}^2

. Substituting

E = 0.16\text{ J}

and

m = 0.50\text{ kg}

yields

0.16 = 0.25 v_{\text{max}}^2

, so

v_{\text{max}}^2 = 0.64

and

v_{\text{max}} = 0.8\text{ m/s}

.

Step-by-Step Solution

1

Relate total energy to maximum kinetic energy

E = \frac{1}{2} m v_{\text{max}}^2

At the equilibrium position, potential energy is zero and total mechanical energy is entirely kinetic.

2

Substitute given values into the equation

0.16 = \frac{1}{2} (0.50) v_{\text{max}}^2

Mass

m = 0.50\text{ kg}

and total energy

E = 0.16\text{ J}

are provided.

3

Solve for the maximum speed

v_{\text{max}} = \sqrt{\frac{2 \times 0.16}{0.50}} = \sqrt{0.64} = 0.8\text{ m/s}

Solving for

v_{\text{max}}

gives

0.8\text{ m/s}

.

Key Concept

Conservation of Energy in Simple Harmonic Motion

Question 1424Question →

An object is placed $24\text{ cm}$ in front of a concave mirror with a focal length of $8\text{ cm}$ . What is the distance of the image from the mirror in $\text{cm}$ ?

Show answer & explanation

Answer: 12

Answer

The distance of the image from the mirror is

12\text{ cm}

.

Applying the spherical mirror formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

with focal length

f = 8\text{ cm}

and object distance

u = 24\text{ cm}

yields

\frac{1}{v} = \frac{1}{8} - \frac{1}{24} = \frac{1}{12}

, giving an image distance of

12\text{ cm}

.

Step-by-Step Solution

1

Identify the given parameters and select the appropriate relation.

Focal length

f = 8\text{ cm}

, object distance

u = 24\text{ cm}

, using the mirror equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

.

The mirror formula relates focal length, object distance, and image distance for spherical mirrors.

2

Substitute the given values into the formula and solve for

\frac{1}{v}

.

\frac{1}{8} = \frac{1}{24} + \frac{1}{v} \implies \frac{1}{v} = \frac{1}{8} - \frac{1}{24} = \frac{3-1}{24} = \frac{2}{24} = \frac{1}{12}

Subtracting

\frac{1}{24}

from

\frac{1}{8}

isolates the reciprocal of the image distance.

3

Invert the result to determine the image distance

v

.

v = 12\text{ cm}

Taking the reciprocal of

\frac{1}{12}

gives the image distance in centimeters.

Key Concept

Mirror Formula for Concave Mirrors

Estimated Time:45s

Question 1425Question →

An electric heater rated at $500\text{ W}$ is used to melt $0.20\text{ kg}$ of ice initially at $0^\circ\text{C}$ and heat the resulting water to $20^\circ\text{C}$ . Assuming no heat losses to the surroundings, what is the time required in seconds? (Specific latent heat of fusion of ice = $3.36 \times 10^5\text{ J kg}^{-1}$ , specific heat capacity of water = $4.2 \times 10^3\text{ J kg}^{-1}\text{ K}^{-1}$ )

Show answer & explanation

Answer: 168

Answer

168 s

The total energy required is the sum of the heat required for melting (

mL_f = 67,200\text{ J}

) and the heat required for warming the liquid (

mc\Delta T = 16,800\text{ J}

), giving

84,000\text{ J}

. Dividing this by the heater power rating of

500\text{ W}

gives

168\text{ s}

.

Step-by-Step Solution

1

Calculate thermal energy needed to change phase from ice to water at

0^\circ\text{C}

Q_1 = m L_f = 0.20\text{ kg} \times 3.36 \times 10^5\text{ J kg}^{-1} = 67,200\text{ J}

Latent heat of fusion accounts for phase change at constant temperature.

2

Calculate thermal energy needed to raise water temperature from

0^\circ\text{C}

to

20^\circ\text{C}

Q_2 = m c \Delta T = 0.20\text{ kg} \times 4.2 \times 10^3\text{ J kg}^{-1}\text{K}^{-1} \times 20\text{ K} = 16,800\text{ J}

Sensible heat increases kinetic energy of liquid water molecules.

3

Sum energy required for both phase change and temperature change

Q_{\text{total}} = Q_1 + Q_2 = 67,200\text{ J} + 16,800\text{ J} = 84,000\text{ J}

Total energy is the sum of sensible heat and latent heat.

4

Calculate heating time using heater power rating

t = \frac{Q_{\text{total}}}{P} = \frac{84,000\text{ J}}{500\text{ W}} = 168\text{ s}

Power is defined as energy transferred per unit time (

P = \frac{Q}{t}

).

Key Concept

Energy balance combining latent heat and sensible heat

Estimated Time:2m 0s

Question 1426Question →

A simple pendulum on Earth ( $g = 10\text{ m/s}^2$ ) completes $50$ full oscillations in $40\text{ s}$ . The pendulum is then transferred to a lunar station where the acceleration due to gravity is $1.6\text{ m/s}^2$ , and its length is reduced by $64\%$ . What is the time taken, in seconds, for this modified pendulum to complete $30$ oscillations on the lunar station?

Show answer & explanation

Answer: 36

Answer

The time taken for the modified pendulum to complete 30 oscillations on the lunar station is 36 s.

The initial period on Earth is

T_1 = \frac{40}{50} = 0.8\text{ s}

. The formula for the period of a simple pendulum is

T = 2\pi\sqrt{\frac{L}{g}}

. When length decreases by

64\%

, the remaining length ratio is

\frac{L_2}{L_1} = 0.36

. The ratio of gravity is

\frac{g_1}{g_2} = \frac{10}{1.6} = 6.25

. Taking the ratio gives

\frac{T_2}{T_1} = \sqrt{0.36 \times 6.25} = \sqrt{2.25} = 1.5

. Thus, the new period is

T_2 = 1.5 \times 0.8\text{ s} = 1.2\text{ s}

. For

30

oscillations, the total time is

t = 30 \times 1.2\text{ s} = 36\text{ s}

.

Step-by-Step Solution

1

Calculate the initial period of oscillation on Earth

T_1 = 0.8\text{ s}

Period

T_1

is total time divided by the number of oscillations:

T_1 = \frac{40\text{ s}}{50} = 0.8\text{ s}

.

2

Set up the ratio for period under altered length and gravitational field

\frac{T_2}{T_1} = 1.5

Using

T = 2\pi \sqrt{\frac{L}{g}}

, we have

\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1} \cdot \frac{g_1}{g_2}} = \sqrt{(1 - 0.64) \cdot \frac{10}{1.6}} = \sqrt{0.36 \cdot 6.25} = \sqrt{2.25} = 1.5

.

3

Determine the new period of oscillation

T_2 = 1.2\text{ s}

T_2 = 1.5 \times T_1 = 1.5 \times 0.8\text{ s} = 1.2\text{ s}

.

4

Calculate the total time required for 30 oscillations

t_2 = 36\text{ s}

Total time

t_2 = N_2 \times T_2 = 30 \times 1.2\text{ s} = 36\text{ s}

.

Key Concept

Period of a simple pendulum and its dependence on length and gravitational acceleration

Question 1427Question →

In a nuclear fusion reaction, a lithium-6 nucleus ( ${^{6}_{3}\text{Li}}$ ) fuses with a deuterium nucleus ( ${^{2}_{1}\text{H}}$ ) to form two alpha particles ( ${^{4}_{2}\text{He}}$ ). Given the atomic masses of ${^{6}_{3}\text{Li}} = 6.0151\text{ u}$ , ${^{2}_{1}\text{H}} = 2.0141\text{ u}$ , and ${^{4}_{2}\text{He}} = 4.0026\text{ u}$ , and taking $1\text{ u} = 931.5\text{ MeV}$ , what is the total energy released in this reaction in $\text{MeV}$ ?

Show answer & explanation

Answer: 22.356

Answer

The total energy released in the fusion reaction is 22.356 MeV.

The total mass before fusion (

{^{6}_{3}\text{Li}} + {^{2}_{1}\text{H}}

) is

8.0292\text{ u}

, while the total mass after fusion (

2 \times {^{4}_{2}\text{He}}

) is

8.0052\text{ u}

. The resulting mass defect is

\Delta m = 0.0240\text{ u}

. Multiplying this mass loss by

931.5\text{ MeV/u}

gives an energy release of

22.356\text{ MeV}

.

Step-by-Step Solution

1

Calculate the total initial mass of the reactants

m_{\text{initial}} = 6.0151\text{ u} + 2.0141\text{ u} = 8.0292\text{ u}

Add the rest masses of the lithium-6 nucleus and the deuterium nucleus together.

2

Calculate the total final mass of the products

m_{\text{final}} = 2 \times 4.0026\text{ u} = 8.0052\text{ u}

The reaction produces two helium-4 nuclei (alpha particles), so multiply the mass of one alpha particle by 2.

3

Determine the mass defect

\Delta m = m_{\text{initial}} - m_{\text{final}} = 8.0292\text{ u} - 8.0052\text{ u} = 0.0240\text{ u}

Subtract the final mass of products from the initial mass of reactants to find the mass lost.

4

Convert the mass defect into energy released

E = \Delta m \times 931.5\text{ MeV/u} = 0.0240 \times 931.5 = 22.356\text{ MeV}

Apply Einstein's mass-energy equivalence principle using the standard conversion factor

1\text{ u} = 931.5\text{ MeV}

.

Key Concept

Mass Defect and Energy Release in Nuclear Fusion Reactions

Question 1428Question →

A magnetometer stationed at a field site records the horizontal component of the Earth's magnetic field as $36\ \mu\text{T}$ and the total magnetic field intensity as $45\ \mu\text{T}$ . What is the magnitude of the vertical component of the Earth's magnetic field, in $\mu\text{T}$ ?

Show answer & explanation

Answer: 27

Answer

The magnitude of the vertical component of the Earth's magnetic field is

27\ \mu\text{T}

.

The total magnetic field strength

B

is the hypotenuse of a right-angled triangle formed by the horizontal component

B_h

and vertical component

B_v

. By applying the Pythagorean relation

B^2 = B_h^2 + B_v^2

, substituting

B = 45\ \mu\text{T}

and

B_h = 36\ \mu\text{T}

yields

B_v = \sqrt{45^2 - 36^2} = \sqrt{729} = 27\ \mu\text{T}

.

Step-by-Step Solution

1

Relate total magnetic intensity to its orthogonal components

B^2 = B_h^2 + B_v^2

The total magnetic field vector of the Earth is the vector sum of mutually perpendicular horizontal and vertical components.

2

Isolate the vertical component variable

B_v = \sqrt{B^2 - B_h^2}

Applying the Pythagorean theorem allows direct calculation of the missing perpendicular side.

3

Substitute given values and compute numerical result

B_v = \sqrt{45^2 - 36^2} = \sqrt{2025 - 1296} = \sqrt{729} = 27\ \mu\text{T}

Evaluates the exact magnitude of the vertical component.

Key Concept

Orthogonal resolution of Earth's magnetic field components

Estimated Time:1m 15s

Question 1429Question →

A double-glazed window of total surface area $1.5\text{ m}^2$ consists of two glass panes, each of thickness $4.0\text{ mm}$ and thermal conductivity $0.80\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ , separated by a stagnant air gap of thickness $2.0\text{ mm}$ with thermal conductivity $0.025\text{ W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$ . If a steady-state temperature difference of $18.0^\circ\text{C}$ is maintained across the window's outer boundary surfaces, what is the rate of heat transfer through the window in watts?

Show answer & explanation

Answer: 300

Answer

The steady-state rate of heat transfer through the double-glazed window is

300\text{ W}

.

Heat conduction through composite layers in series is governed by the total thermal resistance. The thermal resistance per unit area of each layer is

r = \frac{d}{k}

. For two

4.0\text{ mm}

glass panes (

r = 0.005\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

each) and one

2.0\text{ mm}

air gap (

r = 0.080\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

), the total unit resistance is

r_{\text{total}} = 0.090\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

. Multiplying by the area

1.5\text{ m}^2

and temperature difference

18.0^\circ\text{C}

gives

\frac{Q}{t} = \frac{1.5 \times 18.0}{0.090} = 300\text{ W}

.

Step-by-Step Solution

1

Convert layer thicknesses to standard units (meters)

d_{\text{glass}} = 4.0\text{ mm} = 0.004\text{ m}

,

d_{\text{air}} = 2.0\text{ mm} = 0.002\text{ m}

SI units are required for calculations using thermal conductivity in

\text{W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}

.

2

Calculate thermal resistance per unit area for each layer

r_{\text{glass}} = \frac{0.004}{0.80} = 0.005\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

,

r_{\text{air}} = \frac{0.002}{0.025} = 0.080\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

Thermal resistance per unit area is given by

r = \frac{d}{k}

.

3

Sum thermal resistances in series to obtain total unit resistance

r_{\text{total}} = r_{\text{glass1}} + r_{\text{air}} + r_{\text{glass2}} = 0.005 + 0.080 + 0.005 = 0.090\text{ m}^2\cdot\text{K}\cdot\text{W}^{-1}

Heat flows sequentially through all three layers in series.

4

Calculate rate of heat flow across total window area

\frac{Q}{t} = \frac{A \cdot \Delta T}{r_{\text{total}}} = \frac{1.5 \cdot 18.0}{0.090} = 300\text{ W}

Rate of thermal conduction through composite series layers is

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

where

R_{\text{total}} = \frac{r_{\text{total}}}{A}

.

Key Concept

Series thermal conduction through composite layers and thermal resistance

Question 1430Question →

A battery of electromotive force (e.m.f.) $24\,\text{V}$ and internal resistance $2\,\Omega$ is connected across a parallel network consisting of two resistors of resistances $6\,\Omega$ and $12\,\Omega$ . What is the total electrical energy, in Joules, dissipated in the external circuit during an operating time of $5\,\text{minutes}$ ?

Show answer & explanation

Answer: 19200

Answer

The total electrical energy dissipated in the external circuit over 5 minutes is

19,200\,\text{J}

.

To compute the external energy dissipated, first combine the parallel resistors to get an equivalent external resistance of

4\,\Omega

. Adding the

2\,\Omega

internal resistance yields a total circuit resistance of

6\,\Omega

, which draws

4\,\text{A}

of current from the

24\,\text{V}

battery. The power delivered to the external load is

P_{ext} = I^2 R_p = 4^2 \times 4 = 64\,\text{W}

. Multiplying this power by the time duration in seconds (

5 \times 60 = 300\,\text{s}

) yields

19,200\,\text{J}

.

Step-by-Step Solution

1

Calculate the equivalent resistance of the external parallel circuit

R_p = \frac{R_1 \times R_2}{R_1 + R_2} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\,\Omega

The two external resistors are connected in parallel.

2

Calculate the total current supplied by the battery

I = \frac{E}{R_p + r} = \frac{24}{4 + 2} = \frac{24}{6} = 4\,\text{A}

Ohm's law for a complete circuit accounts for both external resistance and internal resistance.

3

Determine the power dissipated exclusively in the external circuit

P_{ext} = I^2 R_p = (4)^2 \times 4 = 16 \times 4 = 64\,\text{W}

Electrical power dissipated across the external load depends on the total current squared times the equivalent external resistance.

4

Convert time from minutes to seconds and calculate total energy dissipated

t = 5 \times 60 = 300\,\text{s}

,

E = P_{ext} \times t = 64 \times 300 = 19,200\,\text{J}

Electrical energy is the product of power in Watts and time in seconds.

Key Concept

Electrical Energy and Power in Circuits with Internal Resistance

Question 1431Question →

A light ray enters the first face of a glass prism surrounded by air. The prism has a refracting angle of $75^\circ$ and a refractive index of $\sqrt{2}$ . Inside the prism, the ray strikes the second refracting face at the critical angle for total internal reflection. Calculate the angle of incidence, in degrees, at the first face.

Show answer & explanation

Answer: 45

Answer

The angle of incidence at the first face is

45^\circ

.

To find the angle of incidence at the first face, first determine the critical angle

C

at the second glass-air surface using

\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}

, which yields

C = 45^\circ

. Since the ray strikes the second face at this critical angle,

r_2 = 45^\circ

. Next, using the geometric relationship for a prism

A = r_1 + r_2

, the angle of refraction at the first surface is

r_1 = A - r_2 = 75^\circ - 45^\circ = 30^\circ

. Finally, applying Snell's law at the first face gives

\sin i = n \sin r_1 = \sqrt{2} \sin 30^\circ = \frac{\sqrt{2}}{2}

. Taking the inverse sine yields

i = 45^\circ

.

Step-by-Step Solution

1

Find the critical angle at the second face

Critical angle

C = 45^\circ

, so

r_2 = 45^\circ

Light travels from glass to air at the critical angle, so

\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}

.

2

Determine the angle of refraction at the first face

r_1 = 30^\circ

The apex angle of a prism satisfies

A = r_1 + r_2

, hence

r_1 = A - r_2 = 75^\circ - 45^\circ = 30^\circ

.

3

Apply Snell's Law at the entry boundary

\sin i = \frac{\sqrt{2}}{2}

Refraction at the first surface gives

\sin i = n \sin r_1 = \sqrt{2} \sin 30^\circ = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2}

.

4

Solve for the incident angle

i

i = 45^\circ

\arcsin\left(\frac{\sqrt{2}}{2}\right) = 45^\circ

.

Key Concept

Refraction through a prism combined with total internal reflection critical angle condition

Question 1432Question →

An electric water pump with an efficiency of $80\%$ lifts $60\text{ kg}$ of water vertically through a height of $10\text{ m}$ in $1\text{ minute}$ . What is the input power of the pump in watts? (Take $g = 10\text{ m s}^{-2}$ )

Show answer & explanation

Answer: 125

Answer

The input power of the pump is

125\text{ W}

.

The total gravitational potential energy gained by

60\text{ kg}

of water lifted

10\text{ m}

is

W = mgh = 60 \times 10 \times 10 = 6000\text{ J}

. Performed over

60\text{ seconds}

, the useful power output is

100\text{ W}

. Accounting for an efficiency of

80\%

(

0.80

), the required input power is

P_{\text{in}} = \frac{100\text{ W}}{0.80} = 125\text{ W}

.

Step-by-Step Solution

1

Calculate the useful work done to lift the water

W = mgh = 60\text{ kg} \times 10\text{ m s}^{-2} \times 10\text{ m} = 6000\text{ J}

The useful work done equals the gravitational potential energy gained by the lifted mass of water.

2

Determine the useful output power of the pump

P_{\text{out}} = \frac{W}{t} = \frac{6000\text{ J}}{60\text{ s}} = 100\text{ W}

Power is the rate at which work is done, and

1\text{ minute}

must be converted to

60\text{ seconds}

.

3

Calculate the required input power using efficiency

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{100\text{ W}}{0.80} = 125\text{ W}

Efficiency is defined as

\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}

, so

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}}

.

Key Concept

Work, Power, and Efficiency of a Pump System

Estimated Time:1m 30s

Question 1433Question →

Element $E$ has a relative atomic mass of $12.011$ and exists as two naturally occurring isotopes: $^{12}\text{E}$ with a natural abundance of $98.9\%$ and $^{A}\text{E}$ with a natural abundance of $1.1\%$ . What is the mass number ( $A$ ) of the second isotope?

Show answer & explanation

Answer: 13

Answer

The mass number of the second isotope is 13.

Relative atomic mass is calculated as the weighted average of the mass numbers of all naturally occurring isotopes. Setting up the equation

12.011 = \frac{(98.9 \times 12) + (1.1 \times A)}{100}

yields

1201.1 = 1186.8 + 1.1A

, which simplifies to

1.1A = 14.3

and gives

A = 13

.

Step-by-Step Solution

1

State the relationship for relative atomic mass based on isotopic abundance.

\text{RAM} = \frac{(\text{Abundance}_1 \times \text{Mass}_1) + (\text{Abundance}_2 \times \text{Mass}_2)}{100}

Relative atomic mass is the weighted average of the atomic masses of naturally occurring isotopes of an element.

2

Substitute the given numerical values into the formula.

12.011 = \frac{(98.9 \times 12) + (1.1 \times A)}{100}

Inserting the known abundances (

98.9\%

and

1.1\%

) and mass number (

12

) sets up an algebraic equation for the unknown mass number

A

.

3

Solve the algebraic equation for

A

.

1201.1 = 1186.8 + 1.1A \implies 1.1A = 14.3 \implies A = 13

Subtracting the contribution of the first isotope and dividing by the abundance of the second isotope gives the integer mass number

13

.

Key Concept

Calculating isotopic mass number from relative atomic mass and fractional abundances

Question 1434Question →

A student records the time taken for a simple pendulum to complete $20$ complete oscillations as $30\text{ s}$ . What is the period of oscillation of the pendulum, in seconds?

Show answer & explanation

Answer: 1.5

Answer

The period of oscillation of the pendulum is

1.5\text{ s}

.

The period of oscillation (

T

) is the time required for one complete cycle. Dividing the total time (

30\text{ s}

) by the number of oscillations (

20

) gives

1.5\text{ s}

.

Step-by-Step Solution

1

Identify the given values for total time and total number of oscillations.

Total time

t = 30\text{ s}

and number of oscillations

N = 20

.

The period is defined as the time taken for a single complete oscillation.

2

Divide the total time by the number of oscillations to calculate the period.

T = \frac{30\text{ s}}{20} = 1.5\text{ s}

.

Applying the period formula

T = \frac{t}{N}

yields the duration of one period.

Key Concept

Period of Oscillation

Estimated Time:45s

Question 1435Question →

A copper calorimeter of mass $0.15\text{ kg}$ contains $0.25\text{ kg}$ of water at $20.0^\circ\text{C}$ . Dry steam at $100.0^\circ\text{C}$ is passed into the water until the final temperature of the mixture reaches $40.0^\circ\text{C}$ . Assuming no heat is lost to the surroundings, what is the mass of steam condensed, in grams?

(Take specific heat capacity of water $= 4200\text{ J kg}^{-1}\text{ K}^{-1}$ , specific heat capacity of copper $= 400\text{ J kg}^{-1}\text{ K}^{-1}$ , and specific latent heat of vaporization of water $= 2.26 \times 10^6\text{ J kg}^{-1}$ )

Show answer & explanation

Answer: 8.84

Answer

8.84 g

By energy conservation, the heat gained by the cold water and copper vessel must equal the total heat released by the condensing steam and the cooling of that condensed water. The heat gained is (0.25 kg × 4200 J/kg·K + 0.15 kg × 400 J/kg·K) × 20.0 K = 22,200 J. The heat lost per kilogram of steam is 2,260,000 J/kg + 4200 J/kg·K × 60.0 K = 2,512,000 J/kg. Dividing 22,200 J by 2,512,000 J/kg gives 0.0088376 kg, which corresponds to 8.84 g.

Step-by-Step Solution

1

Calculate the thermal energy absorbed by the cold water and copper calorimeter

Q_gained = 22,200 J

Both the water and copper calorimeter increase in temperature from 20.0°C to 40.0°C (ΔT = 20.0 K).

2

Express the heat released by mass m_s of steam as it condenses and cools to 40.0°C

Q_lost = m_s * 2,512,000 J

Steam releases latent heat when condensing at 100.0°C (m_s * L_v) and sensible heat when the condensed water cools from 100.0°C to 40.0°C (m_s * c_w * 60.0).

3

Apply the principle of conservation of thermal energy and solve for mass m_s in grams

m_s = 8.84 g

Setting Q_gained equal to Q_lost yields m_s = 22,200 / 2,512,000 = 0.0088376 kg, which equals 8.84 g.

Key Concept

Thermal energy balance involving phase change (latent heat of vaporization) and sensible heat exchange

Question 1436Question →

In domestic cats, coat color is an X-linked trait where the allele for black fur ( $X^B$ ) and the allele for orange fur ( $X^O$ ) are codominant, with heterozygous females ( $X^B X^O$ ) exhibiting a tortoiseshell phenotype. If an orange male cat ( $X^O Y$ ) is mated with a tortoiseshell female cat ( $X^B X^O$ ), what percentage of their male offspring is expected to have orange fur?

Show answer & explanation

Answer: 50

Answer

50%

The female parent (

X^B X^O

) transmits her

X^B

allele or her

X^O

allele to male offspring with equal probability (50% each). Because male offspring inherit the Y chromosome from their father, half of the male offspring will inherit

X^O

and have orange fur (

X^O Y

), giving an expected percentage of 50%.

Step-by-Step Solution

1

Determine the gametes produced by each parent

The male parent (

X^O Y

) produces

X^O

and

Y

gametic types in equal proportion (1:1). The female parent (

X^B X^O

) produces

X^B

and

X^O

gametic types in equal proportion (1:1).

Segregation of chromosomes during meiosis separates alleles into gametes.

2

Determine the genotypes specifically for male offspring

Male offspring inherit the

Y

chromosome from the male parent and one

X

chromosome from the female parent. The possible male genotypes are

X^B Y

(black fur) and

X^O Y

(orange fur).

Sex determination follows the XX-XY system where male sex is determined by inheriting the paternal Y chromosome.

3

Calculate the percentage of orange males among total male offspring

Out of 2 possible male genotypes (

X^B Y

and

X^O Y

), 1 is orange (

X^O Y

). The ratio is 1 out of 2, which equals 50%.

The question asks for the proportion strictly within male offspring.

Key Concept

Sex-linked codominant inheritance and sex determination in XX-XY systems

Question 1437Question →

A ticker-tape timer connected to a power supply operates at a frequency of $50\text{ Hz}$ . A continuous strip of paper tape pulled through the timer records a sequence of dots. Calculate the total time interval, in seconds, between the $1\text{st}$ dot and the $21\text{st}$ dot on the tape.

Show answer & explanation

Answer: 0.4

Answer

0.4 s

On a ticker tape, the time interval between consecutive dots represents one period

T = \frac{1}{f} = \frac{1}{50\text{ Hz}} = 0.02\text{ s}

. The total elapsed time for

N

dots corresponds to

(N - 1)

spaces. For 21 dots, there are 20 spaces, giving a total time of

20 \times 0.02\text{ s} = 0.40\text{ s}

.

Step-by-Step Solution

1

Find the number of spaces between dots

20 spaces

Between

N

dots on a ticker-tape, there are

(N - 1)

time intervals.

2

Calculate the period per space

0.02 s

The period

T

is the reciprocal of the operating frequency

f = 50\text{ Hz}

.

3

Multiply the number of spaces by the period per space

0.4 s

Total time duration is the product of the total number of intervals and the time for one interval.

Key Concept

Calculation of time interval using ticker-tape timer frequency and dot count

Question 1438Question →

In an experiment to determine the density of a solid sphere, the mass of the sphere is measured as $(50.0 \pm 0.5)\text{ g}$ and its radius is measured as $(1.00 \pm 0.02)\text{ cm}$ . What is the maximum percentage error in the calculated density of the sphere?

Show answer & explanation

Answer: 7

Answer

The maximum percentage error in the calculated density of the sphere is

7.0\%

.

Density is related to mass and radius by

\rho = \frac{m}{\frac{4}{3}\pi r^3}

. In error analysis, the maximum fractional error of a calculated quantity is the sum of the fractional errors of its components multiplied by their respective powers. The mass has a percentage error of

\frac{0.5}{50.0} \times 100\% = 1.0\%

, and the radius has a percentage error of

\frac{0.02}{1.00} \times 100\% = 2.0\%

. Multiplying the radius percentage error by 3 gives

6.0\%

, and adding the mass percentage error of

1.0\%

yields a maximum percentage error of

7.0\%

.

Step-by-Step Solution

1

Determine the percentage error in the measurement of mass (

m

).

Percentage error in mass =

\frac{0.5\text{ g}}{50.0\text{ g}} \times 100\% = 1.0\%

.

Relative error multiplied by 100 gives the percentage error of a measurement.

2

Determine the percentage error in the measurement of radius (

r

).

Percentage error in radius =

\frac{0.02\text{ cm}}{1.00\text{ cm}} \times 100\% = 2.0\%

.

Relative error in radius multiplied by 100 gives its percentage error.

3

Apply the error propagation formula for the density of a sphere.

Maximum percentage error in density =

1.0\% + 3(2.0\%) = 7.0\%

.

Density is given by

\rho = \frac{m}{\frac{4}{3}\pi r^3}

. For a formula of the form

X = A^a B^b

, the fractional error propagates as

\frac{\Delta X}{X} = a\frac{\Delta A}{A} + b\frac{\Delta B}{B}

. Here, the exponent of

r

is 3, so its percentage error is multiplied by 3.

Key Concept

Error propagation in fractional powers and derived physical quantities

Question 1439Question →

A sample of ice with a mass of $0.20\text{ kg}$ at $0^\circ\text{C}$ absorbs $67,200\text{ J}$ of thermal energy to melt completely into water at $0^\circ\text{C}$ . What is the specific latent heat of fusion of ice in $\text{J kg}^{-1}$ ?

Show answer & explanation

Answer: 336000

Answer

The specific latent heat of fusion of ice is

336,000\text{ J kg}^{-1}

.

For a change of state from solid to liquid at constant temperature, the thermal energy transferred is related to mass by

Q = m L_f

. Dividing the absorbed energy (

67,200\text{ J}

) by the mass (

0.20\text{ kg}

) yields the specific latent heat of fusion,

336,000\text{ J kg}^{-1}

.

Step-by-Step Solution

1

Identify the given values from the problem statement

Heat energy

Q = 67,200\text{ J}

, Mass

m = 0.20\text{ kg}

These parameters are required to calculate the specific latent heat.

2

Apply the energy formula for phase transition at constant temperature

Q = m L_f

During melting, temperature remains constant at

0^\circ\text{C}

, so heat absorbed depends only on mass and specific latent heat of fusion.

3

Solve for the specific latent heat of fusion

L_f

L_f = \frac{Q}{m} = \frac{67,200}{0.20} = 336,000\text{ J kg}^{-1}

Dividing total energy by mass gives heat required per kilogram.

Key Concept

Specific Latent Heat of Fusion

Question 1440Question →

Consider the reduction of zinc oxide by carbon monoxide:

\text{ZnO}(s) + \text{CO}(g) \rightarrow \text{Zn}(s) + \text{CO}_2(g)

Given the standard enthalpies of formation (

\Delta H_f^\circ

):
-

\Delta H_f^\circ[\text{ZnO}(s)] = -348.0\text{ kJ mol}^{-1}

-

\Delta H_f^\circ[\text{CO}(g)] = -110.5\text{ kJ mol}^{-1}

-

\Delta H_f^\circ[\text{CO}_2(g)] = -393.5\text{ kJ mol}^{-1}

What is the standard enthalpy change of the reaction, $\Delta H^\circ$ , in $\text{kJ mol}^{-1}$ ?

Show answer & explanation

Answer: 65

Answer

The standard enthalpy change of the reaction is +65.0 kJ mol^-1.

According to Hess's Law, the standard enthalpy change of a reaction is calculated by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products. For this reaction,

\Delta H^\circ = [-393.5 + 0] - [-348.0 + (-110.5)] = -393.5 - (-458.5) = +65.0\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Determine the standard enthalpy of formation for zinc element in standard state

\Delta H_f^\circ[\text{Zn}(s)] = 0\text{ kJ mol}^{-1}

By definition, the standard enthalpy of formation of an element in its standard reference state is zero.

2

Apply Hess's Law relationship using enthalpies of formation

\Delta H^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})

Enthalpy change of a reaction equals the total enthalpy of formation of products minus that of reactants.

3

Calculate the sum of formation enthalpies for products

\sum \Delta H_f^\circ(\text{products}) = -393.5 + 0 = -393.5\text{ kJ mol}^{-1}

Products are 1 mole of

\text{CO}_2(g)

and 1 mole of

\text{Zn}(s)

.

4

Calculate the sum of formation enthalpies for reactants

\sum \Delta H_f^\circ(\text{reactants}) = -348.0 + (-110.5) = -458.5\text{ kJ mol}^{-1}

Reactants are 1 mole of

\text{ZnO}(s)

and 1 mole of

\text{CO}(g)

.

5

Subtract reactant total from product total to obtain reaction enthalpy

\Delta H^\circ = -393.5 - (-458.5) = +65.0\text{ kJ mol}^{-1}

Performing the subtraction

-393.5 + 458.5

yields

+65.0\text{ kJ mol}^{-1}

.

Key Concept

Calculating standard enthalpy change of reaction using standard enthalpies of formation via Hess's Law.

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