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Question #42197

Difficulty: 8/10 | 🟢 Free

The Question

A reversible refrigerator operates between a cold body and a heat sink. The cold body is a liquid undergoing a phase change at a constant temperature $T_L = 15 \text{ K}$ . The heat sink has a temperature-dependent heat capacity given by $C(T) = \alpha T^3$ , where $\alpha = 4 \times 10^{-4} \text{ J/K}^4$ . Initially, the heat sink is at a temperature $T_0 = 20 \text{ K}$ . The refrigerator extracts heat from the cold body and rejects it to the heat sink until the sink reaches a temperature $T_f = 40 \text{ K}$ . What is the minimum work required to drive this refrigerator?

Options

A.
$128 \text{ J}$
B.
$120 \text{ J}$
C.
$150 \text{ J}$
D.
$60 \text{ J}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Refrigerator Coefficient of Performance|Entropy and Reversibility|Calorimetry with Heat Loss

Why This Question Is Tricky

Common Mistakes

This question tests the application of the Second Law of Thermodynamics to a system with a variable thermal load.

A common lethal trap is calculating the total heat rejected ( $Q_H = 240 \text{ J}$ ) and then using an arithmetic average temperature for the heat sink ( $T_{\text{avg}} = 30 \text{ K}$ ) to find the work via $W = Q_H(1 - T_L/T_{\text{avg}})$ , which incorrectly yields $120 \text{ J}$ .

Because the heat capacity follows a Debye-like $T^3$ dependence, the temperature weighting is highly non-linear.

Students must integrate the entropy change $dS = dQ/T$ over the exact temperature path to find the reversible heat extracted, enforcing $\Delta S_{\text{universe}} \ge 0$ .

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermodynamics - Class 11 Part 2
Concept	Refrigerator Coefficient of Performance
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free