Question 1

Q1. A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of fuel if its heat o

Accepted Answer

The rate of consumption of fuel is 19.6875 g/min.

Question 2

Q2. A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C, calculate the coefficient of performance.

Accepted Answer

To calculate the coefficient of performance (COP) of a refrigerator, we need the temperatures of the cold reservoir (inside the refrigerator) and the hot reservoir (room temperature) in Kelvin. The formula for the coefficient of performance of a refrigerator is given by:

COP = T_c / (T_h - T_c)

Where:
- T_c is the temperature of the cold reservoir in Kelvin.
- T_h is the temperature of the hot reservoir in Kelvin.

Given:
Temperature inside the refrigerator, t_c = 9°C
Room temperature, t_h = 36°C

First, convert these temperatures to Kelvin:
T_c = t_c + 273.15 = 9 + 273.15 = 282.15 K
T_h = t_h + 273.15 = 36 + 273.15 = 309.15 K

Now, substitute these values into the COP formula:
COP = 282.15 / (309.15 - 282.15)
COP = 282.15 / 27
COP ≈ 10.45

The coefficient of performance of the refrigerator is approximately 10.45.

Question 3

Q3. A steam engine delivers 5.4 × 10⁸ J of work per minute and consumes 3.6 × 10⁹ J of heat per minute from its boiler. Calculate the efficiency of the engine. How much heat is wasted

Accepted Answer

This problem involves calculating the efficiency of a heat engine and the amount of heat wasted, given the work done and heat input.

**Part 1: Calculate the efficiency of the engine**
The efficiency (η) of a heat engine is defined as the ratio of the work done (W) by the engine to the heat absorbed (Q_h) from the hot reservoir (boiler).

Formula for efficiency: η = W / Q_h

Given:
Work delivered per minute, W = 5.4 × 10⁸ J
Heat consumed per minute from boiler, Q_h = 3.6 × 10⁹ J

Substitute the given values into the efficiency formula:
η = (5.4 × 10⁸ J) / (3.6 × 10⁹ J)
η = (5.4 / 3.6) × (10⁸ / 10⁹)
η = 1.5 × 10⁻¹
η = 0.15

To express efficiency as a percentage, multiply by 100:
η = 0.15 × 100% = 15%

The efficiency of the engine is 0.15 or 15%.

**Part 2: Calculate the heat wasted per minute**
According to the First Law of Thermodynamics for a cyclic process (like a heat engine), the heat absorbed from the hot reservoir (Q_h) is equal to the sum of the work done (W) and the heat rejected to the cold reservoir (Q_c, which is the wasted heat).

Formula: Q_h = W + Q_c
Therefore, heat wasted (Q_c) = Q_h - W

Substitute the given values:
Q_c = (3.6 × 10⁹ J) - (5.4 × 10⁸ J)

To perform the subtraction, make the powers of 10 the same:
Q_c = (36 × 10⁸ J) - (5.4 × 10⁸ J)
Q_c = (36 - 5.4) × 10⁸ J
Q_c = 30.6 × 10⁸ J
Q_c = 3.06 × 10⁹ J

The heat wasted per minute is 3.06 × 10⁹ J.

Question 4

Q4. An ideal gas goes from state A to state B via two processes: first an isobaric expansion and then an isochoric process. In the first process, the pressure is 2 × 10⁵ Pa and the vol

Accepted Answer

Let's denote the states as follows:
Initial state: A
Intermediate state: C (after isobaric expansion)
Final state: B (after isochoric process)

Given data:
State A: P_A = 2 × 10⁵ Pa, V_A = 0.02 m³
Process A → C: Isobaric expansion
P_C = P_A = 2 × 10⁵ Pa
V_C = 0.05 m³
Process C → B: Isochoric process
V_B = V_C = 0.05 m³
P_B = 1 × 10⁵ Pa
Change in internal energy, ΔU = 1.5 × 10⁴ J

**(a) Draw the P-V diagram for the combined process.**
The P-V diagram will show two distinct segments:
1.  **A → C (Isobaric Expansion):** A horizontal line segment from (V_A, P_A) to (V_C, P_C). Pressure remains constant at 2 × 10⁵ Pa while volume increases from 0.02 m³ to 0.05 m³.
2.  **C → B (Isochoric Process):** A vertical line segment from (V_C, P_C) to (V_B, P_B). Volume remains constant at 0.05 m³ while pressure decreases from 2 × 10⁵ Pa to 1 × 10⁵ Pa.

**(b) Calculate the total work done by the gas.**
Total work done (W_total) = Work done in process A→C (W_AC) + Work done in process C→B (W_CB).

**Work done in isobaric process A→C (W_AC):**
For an isobaric process, W = PΔV.
W_AC = P_A (V_C - V_A)
W_AC = (2 × 10⁵ Pa) × (0.05 m³ - 0.02 m³)
W_AC = (2 × 10⁵ Pa) × (0.03 m³)
W_AC = 0.06 × 10⁵ J
W_AC = 6 × 10³ J

**Work done in isochoric process C→B (W_CB):**
For an isochoric process, volume remains constant (ΔV = 0), so the work done is zero.
W_CB = 0 J

**Total work done (W_total):**
W_total = W_AC + W_CB
W_total = 6 × 10³ J + 0 J
W_total = 6 × 10³ J

The total work done by the gas is 6 × 10³ J.

**(c) Calculate the total heat absorbed by the gas if the internal energy of the gas changes by 1.5 × 10⁴ J.**
According to the First Law of Thermodynamics:
ΔU = Q - W
Where:
ΔU = Change in internal energy
Q = Total heat absorbed by the gas
W = Total work done by the gas

We are given ΔU = 1.5 × 10⁴ J and we calculated W = 6 × 10³ J.

Rearrange the formula to solve for Q:
Q = ΔU + W

Substitute the values:
Q = (1.5 × 10⁴ J) + (6 × 10³ J)
To add, make the powers of 10 the same:
Q = (1.5 × 10⁴ J) + (0.6 × 10⁴ J)
Q = (1.5 + 0.6) × 10⁴ J
Q = 2.1 × 10⁴ J

The total heat absorbed by the gas is 2.1 × 10⁴ J.

Question 5

Q5. A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C, calculate the coefficient of performance.

Accepted Answer

To calculate the coefficient of performance (COP) of a refrigerator, we need the temperatures of the cold reservoir (inside the refrigerator) and the hot reservoir (room temperature) in Kelvin. The COP of a refrigerator is given by the formula:

COP = T_c / (T_h - T_c)

Where:
T_c = Temperature of the cold reservoir (in Kelvin)
T_h = Temperature of the hot reservoir (in Kelvin)

Given:
Temperature inside the refrigerator, t_c = 9°C
Room temperature, t_h = 36°C

First, convert these temperatures to Kelvin:
T_c = 9 + 273.15 = 282.15 K
T_h = 36 + 273.15 = 309.15 K

Now, substitute these values into the COP formula:
COP = 282.15 / (309.15 - 282.15)
COP = 282.15 / 27
COP ≈ 10.45

The coefficient of performance of the refrigerator is approximately 10.45.

Question 6

Q6. A heat engine operating between 227°C and 77°C absorbs 6 × 10^4 J of heat from the source in each cycle. Calculate the efficiency of the engine. How much heat is rejected to the si

Accepted Answer

This problem involves a heat engine, and we need to calculate its efficiency and the heat rejected to the sink. We assume it's an ideal (Carnot) heat engine for maximum efficiency calculations.

**Part 1: Calculate the efficiency of the engine.**

Given:
Source temperature, t_h = 227°C
Sink temperature, t_c = 77°C
Heat absorbed from the source, Q_h = 6 × 10^4 J

First, convert temperatures to Kelvin:
T_h = 227 + 273.15 = 500.15 K ≈ 500 K
T_c = 77 + 273.15 = 350.15 K ≈ 350 K

The efficiency (η) of a heat engine is given by:
η = 1 - (T_c / T_h)

Substitute the Kelvin temperatures:
η = 1 - (350 K / 500 K)
η = 1 - 0.7
η = 0.3

To express as a percentage:
η = 0.3 × 100% = 30%

The efficiency of the engine is 0.3 or 30%.

**Part 2: Calculate the heat rejected to the sink per cycle.**

The efficiency can also be expressed in terms of heat absorbed (Q_h) and heat rejected (Q_c):
η = 1 - (Q_c / Q_h)

We can use the calculated efficiency and the given heat absorbed:
0.3 = 1 - (Q_c / (6 × 10^4 J))
Q_c / (6 × 10^4 J) = 1 - 0.3
Q_c / (6 × 10^4 J) = 0.7
Q_c = 0.7 × (6 × 10^4 J)
Q_c = 4.2 × 10^4 J

Alternatively, for a Carnot engine, the ratio of heats is equal to the ratio of absolute temperatures:
Q_c / Q_h = T_c / T_h
Q_c = Q_h × (T_c / T_h)
Q_c = (6 × 10^4 J) × (350 K / 500 K)
Q_c = (6 × 10^4 J) × 0.7
Q_c = 4.2 × 10^4 J

The heat rejected to the sink per cycle is 4.2 × 10^4 J.

Question 7

Q7. Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The ent

Accepted Answer

This scenario describes a free expansion of an ideal gas. Let's analyze each part:

### (a) What is the final pressure of the gas in A and B?

Initially, cylinder A contains an ideal gas at standard temperature and pressure (STP). Standard pressure (P_initial) is 1 atm (or 1.013 × 10^5 Pa). Cylinder B is evacuated. The volumes of cylinders A and B are equal. Let V_A be the initial volume of the gas in cylinder A, and V_B be the volume of cylinder B. Since they have equal capacity, V_A = V_B = V.

When the stopcock is opened, the gas expands to fill both cylinders. The final volume (V_final) available to the gas will be V_A + V_B = V + V = 2V.

Since the system is thermally insulated, and the gas expands into a vacuum, no external work is done (W=0). For an ideal gas undergoing free expansion, the internal energy remains constant (ΔU=0), which implies the temperature also remains constant (ΔT=0). This means the process is isothermal.

Applying Boyle's Law for an isothermal process (P_1V_1 = P_2V_2):
P_initial × V_A = P_final × (V_A + V_B)
P_initial × V = P_final × (2V)
P_final = P_initial / 2

If the initial pressure was 1 atm, the final pressure will be 0.5 atm.

### (b) What is the change in internal energy of the gas?

The system is thermally insulated, so no heat is exchanged with the surroundings (Q = 0). The gas expands into a vacuum, meaning there is no external pressure against which the gas does work. Therefore, the work done by the gas (W) is zero.

According to the First Law of Thermodynamics: ΔU = Q - W
Substituting Q = 0 and W = 0:
ΔU = 0 - 0 = 0

Thus, the change in internal energy of the gas is zero.

### (c) What is the change in the temperature of the gas?

For an ideal gas, the internal energy (U) is solely a function of its temperature (T). Since the change in internal energy (ΔU) is zero (as determined in part b), the change in temperature (ΔT) of the ideal gas must also be zero.

ΔU = nC_vΔT
Since ΔU = 0 and n, C_v are non-zero, then ΔT = 0.

Therefore, the temperature of the gas remains constant.

### (d) Do the answers depend on the nature of the gas?

For an ideal gas, the internal energy depends only on temperature, and during free expansion into a vacuum, no work is done and no heat is exchanged. Therefore, ΔU = 0 and ΔT = 0. The final pressure calculation (P_final = P_initial / 2) also relies on the gas behaving ideally and undergoing an isothermal process.

However, for a real gas, there are intermolecular forces. When a real gas expands, work is done against these intermolecular forces, which can lead to a change in internal energy and thus a change in temperature (typically a cooling effect, known as the Joule-Thomson effect, if the initial temperature is below the inversion temperature). So, for a real gas, ΔU would not be zero, and ΔT would not be zero.

Therefore, the answers (specifically regarding ΔU and ΔT) *do depend on the nature of the gas*. The conclusions (ΔU=0, ΔT=0) are strictly valid only for an ideal gas.

### (e) Is the process adiabatic, isothermal, isobaric, or isochoric?

Let's analyze the characteristics of the process:
*   **Adiabatic:** The system is thermally insulated, meaning no heat exchange (Q=0). So, the process is adiabatic.
*   **Isothermal:** As determined in part (c), the temperature of the ideal gas remains constant (ΔT=0). So, the process is isothermal.
*   **Isobaric:** The pressure changes from P_initial to P_final = P_initial / 2. So, it is not isobaric.
*   **Isochoric:** The volume changes from V to 2V. So, it is not isochoric.

Therefore, for an ideal gas, the free expansion process is both **adiabatic** (Q=0) and **isothermal** (ΔT=0). This is a unique characteristic of free expansion of an ideal gas. It is neither isobaric nor isochoric.

Question 8

Q8. An engine takes in 5000 J of heat from a hot reservoir at 127 °C and expels 3000 J of heat to a cold reservoir at 27 °C. What is the efficiency of the engine?

Accepted Answer

The efficiency of the engine is 40%.

Question 9

Q9. A refrigerator is to maintain eatables kept inside at 9 °C. If room temperature is 36 °C, calculate the coefficient of performance. Assume the refrigerator to be a reversible one.

Accepted Answer

The coefficient of performance (COP) of the refrigerator is 10.44.

Question 10

Q10. A steam engine delivers 5.4 × 10^8 J of work per minute and consumes 3.6 × 10^9 J of heat per minute from its boiler. Calculate the efficiency of the engine. How much heat is waste

Accepted Answer

The efficiency of the engine is 15%. The heat wasted per minute is 3.06 × 10^9 J.

Question 11

Q11. An ideal gas is compressed adiabatically to one-half of its initial volume. If the initial pressure is P and the initial temperature is T, find the final pressure and temperature.

Accepted Answer

The final pressure is 2.64P and the final temperature is 1.32T.

Question 12

Q12. Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains an ideal gas at standard temperature and pressure. B is completely evacuated. The ent

Accepted Answer

This scenario describes a free expansion of an ideal gas. Let's analyze each part:

**(a) What is the final pressure of the gas in A and B?**

**Initial State:**
*   Volume of gas in cylinder A = V_A
*   Pressure of gas in cylinder A = P_A (Standard Pressure, P_0)
*   Temperature of gas in cylinder A = T_A (Standard Temperature, T_0)
*   Cylinder B is evacuated.

**Final State:**
*   When the stopcock is opened, the gas expands to fill both cylinders A and B. Since cylinders A and B have equal capacity, the total final volume (V_final) will be V_A + V_B = V_A + V_A = 2V_A.
*   The process is a free expansion into a vacuum. For an ideal gas undergoing free expansion, the internal energy remains constant (as shown in part b), which implies the temperature also remains constant (as shown in part c).
*   Since the temperature (T) remains constant and the number of moles (n) of the gas also remains constant, we can apply Boyle's Law (P_1V_1 = P_2V_2) or the ideal gas law (PV = nRT).

Using Boyle's Law:
P_initial V_initial = P_final V_final
P_A V_A = P_final (2V_A)
P_final = P_A V_A / (2V_A)
P_final = P_A / 2

Therefore, the final pressure of the gas in A and B will be **P/2** (half of the initial pressure).

**(b) What is the change in internal energy of the gas?**

This process is a **free expansion** of an ideal gas. Let's apply the First Law of Thermodynamics: ΔU = Q - W.

*   **Heat exchange (Q):** The entire system is thermally insulated. This means no heat can enter or leave the system. Therefore, Q = 0.
*   **Work done (W):** The gas expands into a vacuum. There is no external pressure against which the gas has to do work. Hence, the work done by the gas is W = 0.

Substituting these values into the First Law of Thermodynamics:
ΔU = 0 - 0
ΔU = 0

Therefore, the change in internal energy of the gas is **zero**.

**(c) What is the change in the temperature of the gas?**

For an ideal gas, the internal energy (U) depends solely on its temperature (T). Since we found in part (b) that the change in internal energy (ΔU) is zero, it implies that the temperature of the ideal gas must also remain constant.

ΔU = 0 ⇒ ΔT = 0

Therefore, the change in the temperature of the gas is **zero**. The final temperature will be equal to the initial temperature, T.

**(d) Do the intermediate and final states of the process lie on its P-V-T surface?**

*   **P-V-T surface:** This surface represents the equilibrium states of a thermodynamic system. Each point on the surface corresponds to a unique equilibrium state defined by specific values of pressure (P), volume (V), and temperature (T).
*   **Final state:** The final state of the gas (P_final = P/2, V_final = 2V, T_final = T) is an equilibrium state. After the expansion, the gas reaches a new uniform pressure and fills the entire volume at the initial temperature. Thus, the **final state lies on the P-V-T surface**.
*   **Intermediate states:** When the stopcock is suddenly opened, the gas rushes into the evacuated cylinder. During this rapid expansion, the gas is not in thermodynamic equilibrium. There will be pressure and temperature gradients within the gas. For example, the pressure near the stopcock will be different from the pressure at the far end of cylinder B. The system does not have well-defined, uniform values of P, V, and T throughout its volume at any given intermediate instant.

Therefore, the **intermediate states of the process do not lie on the P-V-T surface** because they are non-equilibrium states. Only equilibrium states can be represented on the P-V-T surface.

Question 13

Q13. A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C, calculate the coefficient of performance.

Accepted Answer

To maintain eatables at 9°C, this is the cold reservoir temperature (T_c). The room temperature of 36°C is the hot reservoir temperature (T_h). We need to convert these temperatures to Kelvin before calculating the coefficient of performance (COP) for the refrigerator.

Given:
Temperature of cold reservoir, T_c = 9°C = 9 + 273.15 K = 282.15 K
Temperature of hot reservoir, T_h = 36°C = 36 + 273.15 K = 309.15 K

The coefficient of performance (ε) for a refrigerator is given by the formula:
ε = T_c / (T_h - T_c)

Substituting the values:
ε = 282.15 K / (309.15 K - 282.15 K)
ε = 282.15 K / 27 K
ε ≈ 10.45

Therefore, the coefficient of performance of the refrigerator is approximately 10.45.

Question 14

Q14. A heat pump has a coefficient of performance 6.0. If the pump extracts heat from the cold reservoir at 273 K, what is the temperature of the hot reservoir?

Accepted Answer

Given:
Coefficient of performance of the heat pump, ε = 6.0
Temperature of the cold reservoir, T_c = 273 K

We need to find the temperature of the hot reservoir, T_h.

The coefficient of performance (ε) for a heat pump is given by the formula:
ε = T_h / (T_h - T_c)

Substitute the given values into the formula:
6.0 = T_h / (T_h - 273 K)

Now, solve for T_h:
6.0 * (T_h - 273) = T_h
6.0 * T_h - 6.0 * 273 = T_h
6.0 * T_h - 1638 = T_h

Rearrange the terms to isolate T_h:
6.0 * T_h - T_h = 1638
5.0 * T_h = 1638
T_h = 1638 / 5.0
T_h = 327.6 K

Therefore, the temperature of the hot reservoir is 327.6 K.

Question 15

Q15. A Carnot engine takes 300 J of heat from a source at 500 K and rejects 150 J of heat to a sink. The temperature of the sink is:

Accepted Answer

Given:
Heat taken from the source (hot reservoir), Q_h = 300 J
Temperature of the source (hot reservoir), T_h = 500 K
Heat rejected to the sink (cold reservoir), Q_c = 150 J

We need to find the temperature of the sink (cold reservoir), T_c.

For a Carnot engine, the ratio of heat exchanged is equal to the ratio of the absolute temperatures of the reservoirs:
Q_c / Q_h = T_c / T_h

Substitute the given values into the equation:
150 J / 300 J = T_c / 500 K

Simplify the left side:
0.5 = T_c / 500 K

Now, solve for T_c:
T_c = 0.5 * 500 K
T_c = 250 K

Therefore, the temperature of the sink is 250 K.

Question 16

Q16. A Carnot engine takes 2000 J of heat from a reservoir at 500 K, does some work and rejects 1000 J of heat to a sink. What is the temperature of the sink?

Accepted Answer

Given:
Heat absorbed from the hot reservoir, Q_h = 2000 J
Temperature of the hot reservoir, T_h = 500 K
Heat rejected to the cold reservoir, Q_c = 1000 J

We need to find the temperature of the cold reservoir (sink), T_c.

For a Carnot engine, the relationship between the heat exchanged and the absolute temperatures of the reservoirs is:
Q_c / Q_h = T_c / T_h

Substitute the given values into the equation:
1000 J / 2000 J = T_c / 500 K

Simplify the left side:
0.5 = T_c / 500 K

Now, solve for T_c:
T_c = 0.5 * 500 K
T_c = 250 K

Therefore, the temperature of the sink is 250 K.

Question 17

Q17. A Carnot engine works between two temperatures 300 K and 600 K. If it absorbs 1000 J of heat from the source, calculate the work done and the heat rejected to the sink.

Accepted Answer

Given:
Temperature of the cold reservoir (sink), T_c = 300 K
Temperature of the hot reservoir (source), T_h = 600 K
Heat absorbed from the source, Q_h = 1000 J

We need to calculate:
1. Work done (W)
2. Heat rejected to the sink (Q_c)

**Part 1: Calculate the heat rejected to the sink (Q_c)**
For a Carnot engine, the ratio of heat exchanged is equal to the ratio of the absolute temperatures:
Q_c / Q_h = T_c / T_h

Substitute the given values:
Q_c / 1000 J = 300 K / 600 K
Q_c / 1000 J = 0.5
Q_c = 0.5 * 1000 J
Q_c = 500 J

So, the heat rejected to the sink is 500 J.

**Part 2: Calculate the work done (W)**
The work done by a heat engine is the difference between the heat absorbed from the source and the heat rejected to the sink:
W = Q_h - Q_c

Substitute the calculated value of Q_c and the given Q_h:
W = 1000 J - 500 J
W = 500 J

Alternatively, we can first calculate the efficiency (η) of the Carnot engine:
η = 1 - (T_c / T_h)
η = 1 - (300 K / 600 K)
η = 1 - 0.5
η = 0.5 or 50%

The work done is also given by:
W = η * Q_h
W = 0.5 * 1000 J
W = 500 J

Both methods yield the same result.

Therefore, the work done by the engine is 500 J, and the heat rejected to the sink is 500 J.

Chapter 12: Thermodynamics