Welcome to Lecture 97 of our comprehensive JEE Physics Master Series. In this lecture, we establish the foundational principles governing the directionality and feasibility of natural physical processes: the Second Law of Thermodynamics. While the First Law enforces the conservation of energy ($\Delta U = Q – W$), it fails to predict why heat never spontaneously flows from cold to hot bodies or why complete conversion of heat into mechanical work in a cycle is impossible. Here, we analyze the classical formulations of Kelvin-Planck and Clausius, prove their mutual logical equivalence, explore Carnot’s theorem, and deconstruct standard and multi-concept JEE Advanced problems.
Section 1: Comprehensive Concept Capsule
1. The Physical Necessity for the Second Law
The First Law of Thermodynamics is a quantitative statement of energy conservation: for any cyclic process, $\oint (\delta Q – \delta W) = 0$. However, it places no constraint on the direction of energy exchange:
- It does not forbid a stationary block on a rough surface from spontaneously cooling down, converting its internal thermal energy into kinetic energy, and accelerating.
- It does not forbid heat from transferring spontaneously from an ice cube into boiling water.
- It permits $100\%$ conversion of thermal energy extracted from a single heat reservoir into mechanical work, provided $\Delta U = 0$.
Experimental observation reveals that macroscopic spontaneous processes are inherently irreversible and possess a preferred direction. The Second Law provides the directional criterion.
2. The Kelvin-Planck Statement (Heat Engines)
Statement: It is impossible for any device operating on a thermodynamic cycle to receive heat from a single thermal reservoir and produce an equivalent net amount of work.
- Implication: No heat engine can have a thermal efficiency of $\eta = 1$ ($100\%$). A “sink” or low-temperature reservoir is an absolute physical necessity for continuous mechanical work extraction.
- Thermal Efficiency ($\eta$):
$$\eta = \frac{W_{\text{net}}}{Q_{\text{in}}} = \frac{Q_H – Q_C}{Q_H} = 1 – \frac{Q_C}{Q_H}$$
where $Q_H$ is heat absorbed from a hot reservoir at $T_H$, $Q_C$ is heat rejected to a cold reservoir at $T_C$, and $W_{\text{net}} = Q_H – Q_C$. - Perpetual Motion Machine of the Second Kind (PMM-2): A hypothetical cyclic device that converts heat extracted from a single source entirely into mechanical work without rejecting heat to a sink ($Q_C = 0 \implies \eta = 1$). The Kelvin-Planck statement asserts that a PMM-2 cannot exist.
3. The Clausius Statement (Refrigerators & Heat Pumps)
Statement: It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a cooler body to a hotter body.
- Implication: Heat cannot flow spontaneously from a body at lower temperature to a body at higher temperature without external mechanical or electrical work input ($W_{\text{ext}} > 0$).
- Coefficient of Performance (COP):
- Refrigerator ($\beta$): Measures cooling extraction efficiency from the refrigerated space:
$$\beta = \frac{Q_C}{W_{\text{ext}}} = \frac{Q_C}{Q_H – Q_C}$$ - Heat Pump ($\beta_{\text{HP}}$): Measures heating efficiency delivered to the warmer space:
$$\beta_{\text{HP}} = \frac{Q_H}{W_{\text{ext}}} = \frac{Q_H}{Q_H – Q_C} = \beta + 1$$
- Refrigerator ($\beta$): Measures cooling extraction efficiency from the refrigerated space:
4. Formal Proof of Equivalence
The Kelvin-Planck and Clausius statements appear to describe different systems, but they are logically isomorphic: a violation of one statement implies a violation of the other.
Proposition I: Violation of Clausius $\implies$ Violation of Kelvin-Planck
- Assume Clausius is false. Then there exists an ideal refrigerator $R$ that transfers heat $Q_C$ from a cold reservoir at $T_C$ to a hot reservoir at $T_H$ in a complete cycle with zero work input ($W = 0$).
- Coupled to this, operate a standard heat engine $E$ between the same reservoirs. Let engine $E$ extract heat $Q_H$ ($Q_H > Q_C$) from $T_H$, perform work $W = Q_H – Q_C$, and reject heat $Q_C$ to the cold reservoir at $T_C$.
- Now combine refrigerator $R$ and engine $E$ into a single composite system:
- Net heat drawn from the cold reservoir: $Q_C – Q_C = 0$.
- Net heat drawn from the hot reservoir: $Q_H – Q_C$.
- Net work produced by composite system: $W = Q_H – Q_C$.
- The combined system operates in a complete thermodynamic cycle, exchanges heat with only a single thermal reservoir at $T_H$, and produces equivalent net mechanical work $W = Q_H – Q_C$ with zero heat rejection. This is a direct violation of the Kelvin-Planck statement.
Proposition II: Violation of Kelvin-Planck $\implies$ Violation of Clausius
- Assume Kelvin-Planck is false. Then there exists a heat engine $E$ that absorbs heat $Q_H$ from a reservoir at $T_H$ and converts it completely into work $W = Q_H$, rejecting no heat ($Q_C = 0$).
- Use this entire work output $W$ to drive a standard refrigerator $R$ operating between reservoirs $T_C$ and $T_H$. The refrigerator absorbs heat $Q_C$ from $T_C$ and delivers heat $Q_H’ = Q_C + W = Q_C + Q_H$ to the reservoir at $T_H$.
- Evaluate the net effect of the combined engine-refrigerator system:
- Net heat extracted from cold reservoir: $Q_C$.
- Net heat delivered to hot reservoir: $Q_H’ – Q_H = (Q_C + Q_H) – Q_H = Q_C$.
- Net external work input: $0$ (the engine powers the refrigerator internally).
- The composite machine operates in a cycle and produces no effect other than transferring heat $Q_C$ from a colder reservoir to a hotter reservoir with zero net external work. This is a direct violation of the Clausius statement.
5. Carnot’s Theorems & Thermodynamic Temperature Scale
Nicolas Léonard Sadi Carnot established two seminal corollaries derived directly from the Second Law:
- Theorem 1: No heat engine operating between two given constant-temperature thermal reservoirs can be more efficient than a completely reversible (Carnot) engine operating between the same reservoirs:
$$\eta_{\text{irreversible}} \le \eta_{\text{Carnot}}$$ - Theorem 2: All reversible engines operating between the same two thermal reservoirs have identical efficiencies, completely independent of the working substance:
$$\eta_{\text{rev}} = 1 – \frac{T_C}{T_H}$$
6. Clausius Inequality & Definition of Entropy
For any arbitrary closed cyclic process:
$$\oint \frac{\delta Q}{T} \le 0$$
- $\oint \frac{\delta Q}{T} = 0 \iff \text{The cycle is internally reversible.}$
- $\oint \frac{\delta Q}{T} < 0 \iff \text{The cycle is irreversible.}$
- $\oint \frac{\delta Q}{T} > 0 \iff \text{Thermodynamically impossible.}$
Since $\oint_{\text{rev}} \frac{\delta Q}{T} = 0$, the quantity $dS = \left(\frac{\delta Q}{T}\right)_{\text{rev}}$ is an exact differential, defining a state property: Entropy ($S$). For any isolated system, $\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} \ge 0$.
Section 2: Four Graded Solved Examples
Example 1 (JEE Main): Dual Carnot Engines in Series
Problem: Two Carnot engines $A$ and $B$ operate in series. Engine $A$ absorbs heat at temperature $T_1 = 900\text{ K}$ and rejects heat to an intermediate reservoir at temperature $T_2$. Engine $B$ absorbs this heat rejected by engine $A$ and in turn rejects heat to a heat sink at $T_3 = 400\text{ K}$. Determine the intermediate temperature $T_2$ if:
(a) The two engines perform equal amounts of work ($W_A = W_B$).
(b) The two engines have equal thermal efficiencies ($\eta_A = \eta_B$).
Solution:
- Part (a): Equal Work Output ($W_A = W_B$)
For engine $A$: $W_A = Q_1 – Q_2 = Q_1 \left(1 – \frac{T_2}{T_1}\right)$.
For engine $B$: $W_B = Q_2 – Q_3 = Q_2 \left(1 – \frac{T_3}{T_2}\right)$.
For a Carnot engine, $\frac{Q_1}{T_1} = \frac{Q_2}{T_2} = \frac{Q_3}{T_3} = C$.
Therefore:
$$W_A = C(T_1 – T_2) \quad \text{and} \quad W_B = C(T_2 – T_3)$$
Setting $W_A = W_B$:
$$T_1 – T_2 = T_2 – T_3 \implies T_2 = \frac{T_1 + T_3}{2}$$
$$T_2 = \frac{900 + 400}{2} = 650\text{ K}$$ - Part (b): Equal Efficiencies ($\eta_A = \eta_B$)
$$\eta_A = 1 – \frac{T_2}{T_1}, \quad \eta_B = 1 – \frac{T_3}{T_2}$$
Setting $\eta_A = \eta_B$:
$$1 – \frac{T_2}{T_1} = 1 – \frac{T_3}{T_2} \implies \frac{T_2}{T_1} = \frac{T_3}{T_2}$$
$$T_2^2 = T_1 T_3 \implies T_2 = \sqrt{T_1 T_3}$$
$$T_2 = \sqrt{900 \times 400} = \sqrt{360000} = 600\text{ K}$$
Example 2 (JEE Main/Advanced): Real vs. Carnot Refrigerator Dynamics
Problem: A domestic refrigerator maintains the inside temperature at $-13^\circ\text{C}$ while the room temperature is $27^\circ\text{C}$. The actual refrigerator has a coefficient of performance equal to $50\%$ of an ideal Carnot refrigerator. If the heat leaking into the compartment from the room is $130\text{ W}$, compute the electrical power input required to run the compressor.
Solution:
- Convert temperatures to Kelvin:
$$T_C = 273.15 – 13 = 260\text{ K}, \quad T_H = 273.15 + 27 = 300\text{ K}$$ - Calculate ideal Carnot COP ($\beta_{\text{Carnot}}$):
$$\beta_{\text{Carnot}} = \frac{T_C}{T_H – T_C} = \frac{260}{300 – 260} = \frac{260}{40} = 6.5$$ - Actual COP ($\beta_{\text{actual}}$):
$$\beta_{\text{actual}} = 0.50 \times \beta_{\text{Carnot}} = 0.50 \times 6.5 = 3.25 = \frac{13}{4}$$ - Rate of cooling required to maintain steady temperature:
$$\frac{dQ_C}{dt} = 130\text{ J/s} = 130\text{ W}$$ - Electrical power required ($P_{\text{in}}$):
$$\beta_{\text{actual}} = \frac{dQ_C/dt}{P_{\text{in}}} \implies P_{\text{in}} = \frac{dQ_C/dt}{\beta_{\text{actual}}} = \frac{130}{3.25} = 40\text{ W}$$
Example 3 (JEE Advanced): Engine Driving a Refrigerator Between Shared Reservoirs
Problem: A reversible Carnot heat engine operates between reservoirs at temperatures $T_1 = 800\text{ K}$ and $T_2 = 300\text{ K}$. It drives a reversible Carnot refrigerator that extracts heat from a cold storage chamber at $T_3 = 240\text{ K}$ and exhausts heat into the intermediate reservoir at $T_2 = 300\text{ K}$. If the total rate of heat delivered to the reservoir at $T_2$ is $100\text{ kW}$, calculate the rate of heat supplied by the reservoir at $T_1$.
Solution:
- Let $Q_1$ be the heat absorbed per unit time by engine $E$ from reservoir $T_1$, and $Q_2$ be the heat rejected to $T_2$.
For Carnot engine $E$:
$$W = Q_1 \left(1 – \frac{T_2}{T_1}\right) = Q_1 \left(1 – \frac{300}{800}\right) = \frac{5}{8} Q_1$$
$$Q_2 = Q_1 – W = \frac{3}{8} Q_1$$ - The entire work output $W$ drives refrigerator $R$.
For Carnot refrigerator $R$ operating between $T_3 = 240\text{ K}$ and $T_2 = 300\text{ K}$:
$$\beta = \frac{Q_3}{W} = \frac{T_3}{T_2 – T_3} = \frac{240}{300 – 240} = \frac{240}{60} = 4$$
Hence, heat extracted from $T_3$ is:
$$Q_3 = 4 W = 4 \left(\frac{5}{8} Q_1\right) = \frac{5}{2} Q_1$$ - Heat rejected by the refrigerator to the reservoir at $T_2$:
$$Q_2′ = Q_3 + W = \frac{5}{2} Q_1 + \frac{5}{8} Q_1 = \frac{25}{8} Q_1$$ - Total heat rate rejected to the reservoir at $T_2$:
$$Q_{2,\text{total}} = Q_2 + Q_2′ = \frac{3}{8} Q_1 + \frac{25}{8} Q_1 = \frac{28}{8} Q_1 = \frac{7}{2} Q_1$$ - Given $Q_{2,\text{total}} = 100\text{ kW}$:
$$\frac{7}{2} Q_1 = 100 \implies Q_1 = \frac{200}{7}\text{ kW} \approx 28.57\text{ kW}$$
Example 4 (JEE Advanced): Finite Heat Capacities & Maximum Work Extraction
Problem: Two identical finite bodies of constant heat capacity $C$ are initially at absolute temperatures $T_1$ and $T_2$ ($T_1 > T_2$). A reversible heat engine operates between these two bodies in infinitesimal cycles until both bodies reach a common equilibrium temperature $T_f$.
(a) Find the final equilibrium temperature $T_f$.
(b) Calculate the maximum theoretical work $W_{\text{max}}$ produced by the engine.
Solution:
- Part (a): Determining $T_f$ via Reversibility (Clausius Theorem)
During any infinitesimal step, when body 1 is at temperature $T_A$ and body 2 is at $T_B$, the heat extracted from body 1 is $\delta Q_1 = -C\,dT_A$, and heat delivered to body 2 is $\delta Q_2 = C\,dT_B$.
For maximum work extraction, the combined system operates reversibly ($\Delta S_{\text{universe}} = 0$):
$$\Delta S_1 + \Delta S_2 = 0$$
$$\int_{T_1}^{T_f} \frac{C\,dT_A}{T_A} + \int_{T_2}^{T_f} \frac{C\,dT_B}{T_B} = 0$$
$$C \ln\left(\frac{T_f}{T_1}\right) + C \ln\left(\frac{T_f}{T_2}\right) = 0 \implies C \ln\left(\frac{T_f^2}{T_1 T_2}\right) = 0$$
$$\frac{T_f^2}{T_1 T_2} = 1 \implies T_f = \sqrt{T_1 T_2}$$
Notice: Under spontaneous irreversible thermal contact, the final temperature would be the arithmetic mean $\frac{T_1 + T_2}{2}$. For reversible work extraction, it is the geometric mean $\sqrt{T_1 T_2}$. - Part (b): Maximum Work Produced
From the First Law applied to the entire composite system:
$$W_{\text{max}} = -\Delta U_{\text{system}} = -[(\Delta U)_1 + (\Delta U)_2]$$
$$W_{\text{max}} = -\left[ C(T_f – T_1) + C(T_f – T_2) \right] = C(T_1 + T_2 – 2T_f)$$
Substitute $T_f = \sqrt{T_1 T_2}$:
$$W_{\text{max}} = C\left(T_1 + T_2 – 2\sqrt{T_1 T_2}\right) = C\left(\sqrt{T_1} – \sqrt{T_2}\right)^2$$
Section 3: 10-Question Practice Worksheet
Single Choice Questions (JEE Main Standard)
Q1. A Carnot engine operating between temperatures $T_1 = 600\text{ K}$ and $T_2 = 300\text{ K}$ delivers $W = 600\text{ J}$ of work per cycle. The heat absorbed from the high-temperature reservoir per cycle is:
(A) $900\text{ J}$
(B) $1200\text{ J}$
(C) $1500\text{ J}$
(D) $1800\text{ J}$
Q2. An inventor claims to have designed an engine that absorbs $100\text{ kJ}$ of heat from a thermal bath at $400\text{ K}$, delivers $60\text{ kJ}$ of mechanical work, and rejects $40\text{ kJ}$ of heat to a reservoir at $200\text{ K}$. This engine:
(A) Violates the First Law only
(B) Violates the Second Law only
(C) Violates both the First and Second Laws
(D) Violates neither law and is fully realizable
Q3. If the absolute temperature of the source of a Carnot engine is increased while keeping the sink temperature constant, the efficiency of the engine $\eta$:
(A) Increases
(B) Decreases
(C) Remains unchanged
(D) Becomes zero
Q4. A refrigerator has a coefficient of performance $\beta = 5$. If the temperature inside the freezer compartment is $-20^\circ\text{C}$, the temperature of the surroundings to which it exhausts heat must be:
(A) $20^\circ\text{C}$
(B) $30.6^\circ\text{C}$
(C) $40.5^\circ\text{C}$
(D) $50.0^\circ\text{C}$
Q5. Which of the following cyclic integrals represents an irreversible thermodynamic cycle?
(A) $\oint \frac{\delta Q}{T} > 0$
(B) $\oint \frac{\delta Q}{T} = 0$
(C) $\oint \frac{\delta Q}{T} < 0$
(D) $\oint dS < 0$
Multiple Correct Questions (JEE Advanced Standard)
Q6. Which of the following statements are logically equivalent to the Second Law of Thermodynamics?
(A) The efficiency of any heat engine operating between two temperatures cannot exceed $1 – T_C/T_H$.
(B) Heat cannot spontaneously flow from a colder body to a hotter body without external work.
(C) The total entropy of an isolated system can never decrease over time.
(D) Absolute zero temperature can be attained in a finite number of thermodynamic operations.
Q7. A Carnot engine and an irreversible engine operate between the same two thermal reservoirs at temperatures $T_H$ and $T_C$. Which of the following statements must be TRUE?
(A) $\eta_{\text{irrev}} < \eta_{\text{Carnot}}$
(B) For the same heat absorbed $Q_H$, $W_{\text{irrev}} < W_{\text{Carnot}}$
(C) For the same heat absorbed $Q_H$, $Q_{C,\text{irrev}} > Q_{C,\text{Carnot}}$
(D) $\oint \frac{\delta Q}{T} = 0$ for both engines
Q8. For a reversible heat engine operating between three thermal reservoirs at temperatures $T_1 = 600\text{ K}$, $T_2 = 400\text{ K}$, and $T_3 = 300\text{ K}$, which of the following combinations of heat transfers ($Q_1, Q_2, Q_3$, where positive denotes heat absorbed by the engine) are physically permissible?
(A) $Q_1 = +600\text{ J}, Q_2 = -400\text{ J}, Q_3 = -200\text{ J}$
(B) $Q_1 = +300\text{ J}, Q_2 = -200\text{ J}, Q_3 = -100\text{ J}$
(C) $Q_1 = +600\text{ J}, Q_2 = 0\text{ J}, Q_3 = -300\text{ J}$
(D) $Q_1 = +1200\text{ J}, Q_2 = -800\text{ J}, Q_3 = -400\text{ J}$
Numerical Answer Type Questions
Q9. A Carnot engine has an efficiency of $40\%$. When the temperature of its sink is decreased by $50\text{ K}$, its efficiency increases to $50\%$. Determine the temperature of the source (in Kelvin).
Q10. An ideal heat pump is used to heat a classroom maintained at $21^\circ\text{C}$ ($294\text{ K}$) by extracting heat from outside air at $-3^\circ\text{C}$ ($270\text{ K}$). The building loses heat through walls at a constant rate of $24.5\text{ kW}$. Determine the minimum mechanical power input (in $\text{kW}$) required to drive the heat pump.
Section 4: Detailed Step-by-Step Solutions & Quick Answer Key
Quick Answer Key
| Q1 | Q2 | Q3 | Q4 | Q5 | Q6 | Q7 | Q8 | Q9 | Q10 |
|---|---|---|---|---|---|---|---|---|---|
| B | B | A | B | C | A, B, C | A, B, C | C | 500 | 2.0 |
Step-by-Step Solutions
Sol 1.
The efficiency of the Carnot engine is:
$$\eta = 1 – \frac{T_2}{T_1} = 1 – \frac{300}{600} = 0.50$$
Since $\eta = \frac{W}{Q_1}$:
$$Q_1 = \frac{W}{\eta} = \frac{600\text{ J}}{0.50} = 1200\text{ J}$$
Correct Option: (B)
Sol 2.
Check First Law: $Q_{\text{net}} = Q_{\text{in}} – Q_{\text{out}} = 100 – 40 = 60\text{ kJ} = W$. The First Law is satisfied.
Check Second Law: Maximum theoretical Carnot efficiency between these reservoirs is:
$$\eta_{\text{max}} = 1 – \frac{T_C}{T_H} = 1 – \frac{200}{400} = 0.50 = 50\%$$
Claimed engine efficiency:
$$\eta_{\text{claimed}} = \frac{W}{Q_{\text{in}}} = \frac{60}{100} = 0.60 = 60\%$$
Since $\eta_{\text{claimed}} > \eta_{\text{max}}$, it violates Carnot’s theorem and the Second Law.
Correct Option: (B)
Sol 3.
$$\eta = 1 – \frac{T_C}{T_H}$$
Differentiating with respect to $T_H$:
$$\frac{\partial \eta}{\partial T_H} = \frac{T_C}{T_H^2} > 0$$
Hence, increasing the source temperature $T_H$ increases the thermodynamic efficiency.
Correct Option: (A)
Sol 4.
Inside temperature $T_C = -20 + 273.15 = 253.15\text{ K} \approx 253\text{ K}$.
For an ideal refrigerator:
$$\beta = \frac{T_C}{T_H – T_C} \implies 5 = \frac{253}{T_H – 253}$$
$$5(T_H – 253) = 253 \implies 5T_H – 1265 = 253$$
$$5T_H = 1518 \implies T_H = 303.6\text{ K}$$
In Celsius: $t_H = 303.6 – 273 = 30.6^\circ\text{C}$.
Correct Option: (B)
Sol 5.
According to Clausius’ Inequality:
– For a reversible cycle: $\oint \frac{\delta Q}{T} = 0$.
– For an irreversible cycle: $\oint \frac{\delta Q}{T} < 0$.
– $\oint \frac{\delta Q}{T} > 0$ is impossible.
Note that entropy $S$ is a state function, so $\oint dS = 0$ for any cycle.
Correct Option: (C)
Sol 6.
Statements (A), (B), and (C) are formulations and direct mathematical corollaries of the Second Law of Thermodynamics (Carnot’s theorem, Clausius statement, and Principle of Entropy Increase).
Statement (D) relates to the Third Law of Thermodynamics (Nernst Heat Theorem / Unattainability of absolute zero).
Correct Options: (A, B, C)
Sol 7.
By Carnot’s theorem:
– (A) is True: An irreversible engine is strictly less efficient than a reversible one between the same boundaries ($\eta_{\text{irrev}} < \eta_{\text{Carnot}}$).
– (B) is True: Since $W = \eta Q_H$ and $Q_H$ is identical, $W_{\text{irrev}} < W_{\text{Carnot}}$.
– (C) is True: From energy conservation, $Q_C = Q_H – W$. Since $W_{\text{irrev}} < W_{\text{Carnot}}$, it rejects more heat: $Q_{C,\text{irrev}} > Q_{C,\text{Carnot}}$.
– (D) is False: For an irreversible cycle, $\oint \frac{\delta Q}{T} < 0$, while for Carnot $\oint \frac{\delta Q}{T} = 0$.
Correct Options: (A, B, C)
Sol 8.
For a reversible cycle involving multiple reservoirs, Clausius’ theorem requires:
$$\sum \frac{Q_i}{T_i} = 0 \implies \frac{Q_1}{600} + \frac{Q_2}{400} + \frac{Q_3}{300} = 0$$
Let us test the options:
– For (A): $\frac{600}{600} + \frac{-400}{400} + \frac{-200}{300} = 1 – 1 – \frac{2}{3} = -\frac{2}{3} \neq 0$.
– For (B): $\frac{300}{600} + \frac{-200}{400} + \frac{-100}{300} = \frac{1}{2} – \frac{1}{2} – \frac{1}{3} = -\frac{1}{3} \neq 0$.
– For (C): $\frac{600}{600} + \frac{0}{400} + \frac{-300}{300} = 1 + 0 – 1 = 0$. This strictly satisfies the reversibility condition.
– For (D): $\frac{1200}{600} + \frac{-800}{400} + \frac{-400}{300} = 2 – 2 – \frac{4}{3} = -\frac{4}{3} \neq 0$.
Correct Option: (C)
Sol 9.
Initial state:
$$\eta_1 = 1 – \frac{T_2}{T_1} = 0.40 \implies \frac{T_2}{T_1} = 0.60 \implies T_2 = 0.60 T_1$$
When sink temperature is reduced by $50\text{ K}$ ($T_2′ = T_2 – 50$):
$$\eta_2 = 1 – \frac{T_2 – 50}{T_1} = 0.50 \implies \frac{T_2 – 50}{T_1} = 0.50$$
$$T_2 – 50 = 0.50 T_1$$
Substitute $T_2 = 0.60 T_1$:
$$0.60 T_1 – 50 = 0.50 T_1 \implies 0.10 T_1 = 50 \implies T_1 = \frac{50}{0.10} = 500\text{ K}$$
Sol 10.
For an ideal Carnot heat pump delivering heat $Q_H$ to the room at $T_H = 294\text{ K}$ from the cold outside air at $T_C = 270\text{ K}$:
$$\text{COP}_{\text{HP}} = \frac{Q_H}{W} = \frac{T_H}{T_H – T_C} = \frac{294}{294 – 270} = \frac{294}{24} = 12.25$$
To balance heat loss, the pump must deliver $\frac{dQ_H}{dt} = 24.5\text{ kW}$.
$$\text{COP}_{\text{HP}} = \frac{dQ_H/dt}{P_{\text{in}}} \implies P_{\text{in}} = \frac{24.5\text{ kW}}{12.25} = 2.0\text{ kW}$$
The minimum electrical power input required is 2.0 kW.