Unit-10 Thermodynamics – Complete Exercise Solutions | Class 11 Physics | FBISE

Unit-10 Thermodynamics – Complete Exercise Solutions | Class 11 Physics | KPK Text Board, Peshawar | FBISE

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Click on the link below for complete Class 11 Physics notes, MCQs, Q&A, and Numericals.

Class 11 Physics – Unit-wise MCQs & Study Resources (Main-Page)

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This post provides a complete and fully solved exercise of Unit-10 Thermodynamic for Class 11 Physics, strictly according to the KPK Textbook Board / FBISE syllabus. It includes all Exercise MCQs, short-(conceptual) answer questions, comprehensive (long) questions, and numerical problems, solved step-by-step using standard board-style notation, formulas, and clear explanations.

Whether you are preparing for annual board examinations, chapter tests, or entry tests, this guide is designed to be concept-clear, exam-oriented, and high-scoring, helping you master the chapter with confidence.

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A. Choose the best possible answer of the following MCQs.

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1. Assume we can change the equilibrium state of a system via two different processes. The initial and final states are the same. Which of the quantities ΔU, ΔQ, ΔW, and ΔT must be the same for the two processes?

a) Only ΔQ and ΔW
b) Only ΔU and ΔT
c) Only ΔQ and ΔT
d) Only ΔU and ΔW

Correct Answer is option b. Only ΔU and ΔT

Explanation: Internal energy and temperature are state functions, so they depend only on initial and final states. Heat and work depend on the path.

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2. In any process, the maximum amount of mechanical energy that can be converted into heat:

a) Depends upon the amount of friction
b) Depends upon the intake and exhaust temperature
c) Depends upon whether kinetic or potential energy is involved
d) Is 100%

Correct Answer is option d. Is 100%

Explanation: Mechanical energy can be completely converted into heat, but heat cannot be fully converted into work.

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3. In an isothermal change, internal energy:

a) Decreases
b) Increases
c) Becomes zero
d) Remains constant

Correct Answer is option d. Remains constant

Explanation: For an ideal gas, internal energy depends only on temperature. In an isothermal process, temperature remains constant.

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4. A thermos bottle containing hot coffee is vigorously shaken. Consider coffee as the system. Its temperature:

a) Increases
b) Decreases below 0°C
c) Remains the same
d) Decreases

Correct Answer is option a. Increases

Explanation: Mechanical work done on the system increases its internal energy, raising the temperature.

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5. Maximum work can be obtained in the process called:

a) Cyclic
b) Isothermal
c) Adiabatic
d) Isochoric

Correct Answer is option b. Isothermal

Explanation: In reversible isothermal expansion of an ideal gas, maximum work is obtained.

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6. A heat engine takes in 800 J of heat at 1000 K and exhausts 600 J at 400 K. What is the actual efficiency of this engine?

a) 25%
b) 40%
c) 50%
d) 75%

Correct Answer is option a. 25%

Explanation:
Efficiency = (Qin − Qout) / Qin = (800 − 600)/800 = 200/800 = 0.25 = 25%.

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7. If the temperature of the heat source is increased, the efficiency of a Carnot engine:

a) Increases
b) Decreases
c) Remains constant
d) First increases then becomes constant

Correct Answer is option a. Increases

Explanation: Carnot efficiency η = 1 − Tc/Th. Increasing Th increases efficiency.

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8. The triple point of water is:

a) 273.16°F
b) 372.16 K
c) 273.16°C
d) 273.16 K

Correct Answer is option d. 273.16 K

Explanation: The triple point of water occurs at 273.16 K and a pressure of 611 Pa.

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9. A real gas can be approximated to an ideal gas at:

a) Low density
b) High pressure
c) High density
d) Low temperature

Correct Answer is option a. Low density

Explanation: At low density (low pressure and high temperature), intermolecular forces become negligible.

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10. If the volume of a gas is increased four times, then:

a) Temperature and pressure must be doubled
b) At constant pressure, temperature must be increased four times
c) At constant temperature, pressure must be increased four times
d) It cannot be increased

Correct Answer is option b. At constant pressure, temperature must be increased four times

Explanation: According to Charles’ law, V ∝ T at constant pressure.

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11. In which of the following systems is entropy decreasing?

a) A gas is cooled
b) A plate is shattered
c) An egg is scrambled
d) A drop of dye diffuses in water

Correct Answer is option a. A gas is cooled

Explanation: Cooling reduces molecular disorder, so entropy decreases.

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12. If the temperature of the source and sink of a Carnot engine is each decreased by 100 K, the efficiency:

a) Remains constant
b) Becomes 1
c) Increases
d) Decreases

Correct Answer is option d. Decreases

Explanation: Efficiency depends on the ratio Tc/Th. Decreasing both temperatures generally reduces efficiency.

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CONCEPTUAL QUESTIONS

Give short response to the following questions

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1. Why is the Earth not in thermal equilibrium with the Sun?

Answer:
The Earth is not in thermal equilibrium with the Sun because the Sun is at a very high temperature while the Earth is much cooler. Heat continuously flows from the Sun to the Earth, so their temperatures are not equal.

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2. When a block with a hole in it is heated, why does the material around the hole not expand into the hole and make it smaller?

Answer:
When the block is heated, all dimensions expand uniformly. The hole also expands as if it were made of the same material. Therefore, the hole becomes larger, not smaller.

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3. A thermometer is placed in direct sunlight. Will it read the temperature of the air, the Sun, or something else?

Answer:
The thermometer will read its own temperature, which depends on heat gained from sunlight and the surrounding air. It does not measure the Sun’s temperature directly.

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4. The pressure in a gas cylinder containing hydrogen leaks more quickly than oxygen. Why?

Answer:
Hydrogen molecules are much smaller and lighter than oxygen molecules. They diffuse faster through tiny pores and leaks, so hydrogen escapes more quickly.

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5. What happens to the temperature of a room in which an air conditioner is left running on a table in the middle of the room?

Answer:
The room temperature will increase. The air conditioner releases more heat into the room than it removes because of electrical energy consumption and inefficiency.

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6. Why does the pressure of air in an automobile tyre increase when the automobile is driven for a while?

Answer:
Driving causes friction between the tyre and road, increasing the temperature of the air inside the tyre. According to the gas laws, pressure increases with temperature.

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7. On removing the valve, the air escaping from a cycle tube cools. Why? Write the limitations of the first law of thermodynamics.

Answer (Cooling):
The escaping air expands rapidly and does work. Its internal energy decreases, causing a drop in temperature (adiabatic expansion).

Limitations of First Law:

• It does not explain the direction of heat flow.
• It does not explain irreversibility of natural processes.
• It does not set a limit on heat engine efficiency.
• It does not introduce the concept of entropy.

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8. Is it possible, according to the second law of thermodynamics, to construct a heat engine free from thermal pollution?

Answer:
No. According to the second law, no heat engine can convert all heat into work. Some heat must be rejected to the surroundings, causing thermal pollution.

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9. Can specific heat of a gas be zero, infinity, or negative?

Answer:

• Specific heat cannot be zero or negative under normal physical conditions.
• It can become very large (approach infinity) during phase transitions, but not truly infinite for gases in normal conditions.

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10. An inventor claims to have developed a heat engine working between 27°C and 227°C with efficiency 45%. Is the claim valid? Why?

Answer:

No, the claim is invalid.
Maximum efficiency is given by Carnot’s formula:

$$\eta = 1 - \frac{T_c}{T_h}$$

Convert to Kelvin:
Th = 227 + 273 = 500 K
Tc = 27 + 273 = 300 K

$$\eta = 1 - \frac{300}{500} = 0.4 = 40\%$$

So maximum possible efficiency is 40%, not 45%.

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COMPREHENSIVE QUESTIONS

Give extended response to the following question

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1. Explain briefly the terms: System, Surroundings, Boundary, and State Variables.

In thermodynamics, a system is a specified part of the universe chosen for study, such as a gas in a cylinder or water in a container. Everything outside the system that can interact with it is called the surroundings. The boundary is the real or imaginary surface that separates the system from its surroundings and can be fixed or movable. State variables (or state functions) are physical quantities that describe the state of a system, such as pressure, volume, temperature, and internal energy. These variables depend only on the state of the system and not on how the system reached that state.

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2. Distinguish among the three forms of energy: Work, Heat, and Internal Energy.

Work is the energy transferred when a force causes displacement, such as when a gas expands and pushes a piston. Heat is the energy transferred between a system and its surroundings due to a temperature difference. Internal energy is the total microscopic energy of a system, including the kinetic and potential energies of its molecules. Work and heat are modes of energy transfer, while internal energy is a property of the system.

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3. State and explain the First Law of Thermodynamics.

The First Law of Thermodynamics states that energy can neither be created nor destroyed, but it can be converted from one form to another. Mathematically, it is written as:

$$$$ΔU=Q−W 

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. This law expresses the principle of conservation of energy. If heat is supplied to a system, it either increases the internal energy or is used to do work, or both.

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4. Define molar heat capacities Cp and Cv and show that Cp − Cv = R.

The molar heat capacity at constant volume (Cv) is the amount of heat required to raise the temperature of one mole of gas by one kelvin at constant volume. The molar heat capacity at constant pressure (Cp) is the heat required to raise the temperature of one mole of gas by one kelvin at constant pressure.

For an ideal gas, when heated at constant pressure, part of the heat increases internal energy and part is used to do work. Using the first law and ideal gas equation, it can be shown that:

$$$$Cp - Cv = R

where R is the universal gas constant. This relation is known as Mayer’s relation.

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5. Explain reversible and irreversible processes with examples.

A reversible process is a process that can be reversed without leaving any change in the system and surroundings. It occurs infinitely slowly and without friction, such as slow isothermal expansion of gas in a frictionless piston. An irreversible process is one that cannot be reversed without leaving changes in the surroundings. Real processes like rapid expansion, friction, heat flow through finite temperature difference, and mixing of gases are irreversible.

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6. What is meant by a heat engine? What is its purpose and efficiency?

A heat engine is a device that converts heat energy into mechanical work. It operates between a hot reservoir and a cold reservoir. Its main purpose is to extract work from heat energy. The efficiency of a heat engine is defined as the ratio of work output to heat absorbed from the hot reservoir:

$$\eta = \frac{W}{Q_h} = 1 - \frac{Q_c}{Q_h}$$

where Qh is heat absorbed and Qc is heat rejected.

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7. State the second law of thermodynamics and discuss assertions of the first and second laws. Also mention Carnot’s motivation.

The Second Law of Thermodynamics has two main statements:

1. Kelvin-Planck statement: It is impossible to construct a heat engine that converts all absorbed heat into work without rejecting some heat.

2. Clausius statement: Heat cannot flow spontaneously from a colder body to a hotter body without external work.

The First Law states that energy is conserved, but it does not specify the direction of energy transfer. The Second Law introduces the concept of direction and irreversibility of natural processes.

Carnot was motivated by the question: What is the maximum efficiency possible for a heat engine? He developed the ideal Carnot engine to study the theoretical limits of efficiency.

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8. State Carnot’s Theorem about Carnot engine.

Carnot’s Theorem states that no heat engine operating between two given reservoirs can be more efficient than a Carnot engine operating between the same reservoirs, and all reversible engines operating between the same reservoirs have the same efficiency. The efficiency of a Carnot engine depends only on the temperatures of the hot and cold reservoirs and is given by:

$$\eta = 1 - \frac{T_c}{T_h}$$

9. What is a refrigerator? How does it function? Derive expression for coefficient of performance.

A refrigerator is a heat engine operating in reverse. It removes heat from a low-temperature reservoir and rejects it to a high-temperature reservoir by using external work. The Coefficient of Performance (COP) is defined as the ratio of heat extracted from the cold reservoir to the work done:

$$COP = \frac{Q_c}{W}$$

Since W = Qh − Qc, the expression becomes:

$$COP = \frac{Q_c}{Q_h - Q_c}$$

For an ideal refrigerator operating between temperatures Tc and Th:

$$COP = \frac{T_c}{T_h - T_c}$$

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10. Explain entropy and its properties. Express the second law in terms of entropy.

Entropy is a measure of disorder or randomness in a system. It is defined as:

$$dS = \frac{dQ}{T}$$

for a reversible process.

Major properties of entropy include:
Entropy increases in irreversible processes, remains constant in reversible adiabatic processes, and is a state function depending only on the state of the system.

The Second Law in terms of entropy states that the entropy of the universe always increases in natural processes and remains constant for reversible processes. This means energy becomes less useful over time, and natural processes tend toward disorder.

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4. Numerical Problems:

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Numerical 1

Problem: Water at 20 °C falls from a height of 854 meters. If the whole energy is used in increasing the temperature, find out the final temperature. Specific heat of water is 4200 J·kg⁻¹·K⁻¹.

Given:

$T_i = 20\,°\text{C}, h = 854\,\text{m}, c = 4200\,\text{J/kg·K}, g = 9.8\,\text{m/s²}$

To Find: $T_f$

Solution:

Potential energy per kg: $PE = m g h$ 
Assuming all energy converts to heat: $mc\Delta T = m g h$

$$\Delta T = \frac{gh}{c} = \frac{9.8 \times 854}{4200} \approx 1.99\,\text{K} \approx 2\,\text{K}$$$$T_f = T_i + \Delta T = 20 + 2 = 22\,°\text{C}$$

Answer: $T_f\approx 22\mathrm{°}\text{C}$

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Numerical 2

Problem: 25200 J of heat is supplied to the system while the system does 6000 J of work. Calculate the change in internal energy of the system.

Given:
$Q=25200\text{J},W=6000\text{J}$

To Find: $\Delta U$

Solution:
First Law:  $\Delta U=Q-W=25200-6000=19200\text{J}$

Answer:  $\Delta U=19200\text{J}$

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Numerical 3

Problem: A sample of ideal gas is uniformly heated at constant pressure. If 180 J of heat is supplied to the gas, calculate the change in internal energy of the gas and work done by the gas. Take γ = 1.41.

Given:
 Q=180J,γ=1.41

To Find: $\Delta U,W$

Solution:

$$W = \frac{\gamma - 1}{\gamma} Q = \frac{1.41 - 1}{1.41} \cdot 180 \approx 52.5\,\text{J}$$$$\Delta U = Q - W = 180 - 52.5 = 127.5\,\text{J}$$

Answer: $\Delta U\approx 127.5\text{J},W\approx 52.5\text{J}$

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Numerical 4

Problem: Find the efficiency of a Carnot heat engine working between the steam and ice points.

Given:
$T_h = 373\,\text{K}, T_c = 273\,\text{K}$ 

To Find: $\eta$  

Solution:

$$\eta = 1 - \frac{T_c}{T_h} = 1 - \frac{273}{373} = 0.269$$

Answer: $\eta \approx 26.9\%$ 

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Numerical 5

Problem: A Carnot heat engine absorbs 2000 J of heat from the source at 227 °C and rejects 1200 J of heat during each cycle to the sink. Calculate the efficiency of the engine, temperature of the sink, and amount of work done per cycle.

Given:

$Q_h = 2000\,\text{J}, Q_c = 1200\,\text{J}, T_h = 227\,°\text{C} = 500\,\text{K}$

To Find: $\eta, W, T_c$

Solution:

Work done: $W=Q_h-Q_c=2000-1200=800\text{J}\text{<br>}$ 

Efficiency: $\eta = \frac{W}{Q_h} = \frac{800}{2000} = 0.4 = 40\%$ 

Sink temperature: $T_c = T_h (1 - \eta) = 500 \cdot (1 - 0.4) = 300\,\text{K}$ 

Answer: $W = 800\,\text{J}, \eta = 40\%, T_c = 300\,\text{K}$

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Numerical 6

Problem: What is the least amount of work required to freeze 1 g of water at 0 °C using a refrigerator? The temperature of the surrounding is 37 °C. How much heat is passed to the surrounding?

Given:

$m = 1\,\text{g}, L_f = 334\,\text{J/g}, T_c = 273\,\text{K}, T_s = 310\,\text{K}$ 

To Find: $W, Q_h$

Solution:

Heat removed: $Q_c = m L_f = 334\,\text{J}$ 

Coefficient of Performance: 

$COP = \frac{T_c}{T_s - T_c} = \frac{273}{310 - 273} \approx 7.38$ 

Work done: $W = \frac{Q_c}{COP} = 334/7.38 \approx 45.3\,\text{J}$ 

Heat delivered to surroundings: $Q_h = Q_c + W = 334 + 45.3 \approx 379.3\,\text{J}$

Answer: $W \approx 45.3\,\text{J}, Q_h \approx 379.3\,\text{J}$ 

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Numerical 7

Problem: Calculate the change in entropy when 10 kg of water is heated from 90 °C to 100 °C. Take specific heat $c=4180$ J·kg⁻¹·K⁻¹

Given:

$m = 10\,\text{kg}, c = 4180\,\text{J/kg·K}, T_1 = 363\,\text{K}, T_2 = 373\,\text{K}$

To Find: $\Delta S$ 

Solution:

$$\Delta S = m c \ln \frac{T_2}{T_1} = 10 \cdot 4180 \cdot \ln \frac{373}{363}$$$$\ln \frac{373}{363} = \ln 1.0275 \approx 0.0271$$$$\Delta S \approx 10 \cdot 4180 \cdot 0.0271 \approx 1131\,\text{J/K}$$

Answer: $\Delta S \approx 1131\,\text{J/K}$ 

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Numerical 8

Problem: A system absorbs 1176 J of heat while doing 352.8 J of work. Find the change in internal energy. Also, find ΔU when 1050 J heat is absorbed and 84 J work is done. What is ΔU if 210 J heat is removed at constant volume?

Given:
Case 1: $Q = 1176\,\text{J}, W = 352.8\,\text{J}$ 
Case 2: $Q = 1050\,\text{J}, W = 84\,\text{J}$ 
Case 3: $Q = -210\,\text{J}, W = 0$ 

To Find: ΔU

Solution:

$$\Delta U = Q - W$$

Case 1: $1176 - 352.8 = 823.2\,\text{J}$ 

Case 2: $1050 - 84 = 966\,\text{J}$ 

Case 3: $-210 - 0 = -210\,\text{J}$ 

Answer: $\Delta U = 823.2\,\text{J}, 966\,\text{J}, -210\,\text{J}$ 

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Numerical 9

Problem: An ideal gas at 20 °C and 1.50×10⁵ Pa occupies 1.00 L. (a) Find the number of moles. (b) If it expands to twice its volume and pressure drops to 1 atm, find the final temperature.

Given:
$P = 1.5 \times 10^5\,\text{Pa}, V = 1.00 \times 10^{-3}\,\text{m³}, T = 293\,\text{K}, R = 8.314\,\text{J/mol·K}, P_2 = 1.013 \times 10^5\,\text{Pa}, V_2 = 2V$

To Find: $n, T_2$

Solution:

(a) $n = \frac{PV}{RT} = \frac{1.5 \times 10^5 \cdot 1 \times 10^{-3}}{8.314 \cdot 293} \approx 0.0617\,\text{mol}$

(b) $T_2 = \frac{P_2 V_2}{n R} = \frac{1.013 \times 10^5 \cdot 2 \times 10^{-3}}{0.0617 \cdot 8.314} \approx 395\,\text{K}$

 Answer: $n \approx 0.0617\,\text{mol}, T_2 \approx 395\,\text{K}$ 

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Numerical 10

Problem: A block of ice at 273 K is put in thermal contact with steam at 373 K, converting 25 g ice to water at 273 K while condensing some steam at 373 K. Find (a) ΔS of ice, (b) ΔS of steam, (c) ΔS of Universe.

Given:
$m_i = 25\,\text{g}, L_f = 334\,\text{J/g}, L_v = 2260\,\text{J/g}, T_i = 273\,\text{K}, T_s = 373\,\text{K}$

To Find: $\Delta S_i, \Delta S_s, \Delta S_{universe}$

Solution:

(a) $\Delta S_i = \frac{Q}{T} = \frac{25 \cdot 334}{273} \approx 30.6\,\text{J/K}$ 

(b) $Q_{steam}=m_i{L}_f=25\cdot 334=8350\text{J}\text{<br>}\text{<br>}\text{<br>}$ 

Mass of steam condensed: $m_s = \frac{Q_{steam}}{L_v} = \frac{8350}{2260} \approx 3.7\,\text{g}$

$$\Delta S_s = - \frac{m_s L_v}{T_s} = -\frac{3.7 \cdot 2260}{373} \approx -22.4\,\text{J/K}$$

(c) $\Delta S_{universe} = \Delta S_i + \Delta S_s = 30.6 - 22.4 \approx 8.2\,\text{J/K}$

Answer: $\Delta S_i \approx 30.6\,\text{J/K}, \Delta S_s \approx -22.4\,\text{J/K}, \Delta S_{universe} \approx 8.2\,\text{J/K}$

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Click on the link below for complete Class 11 Physics notes, MCQs, Q&A, and Numericals.

Class 11 Physics – Unit-wise MCQs & Study Resources (Main-Page)

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