Question

A metal rod is in thermal contact with the two heat reservoirs both at constant temperature, one at 100K and the other at 200K. The rod conducts 1000 J of heat from the warmer to the colder reservoir. If no energy is exchanged with the surroundings, what is the total change of entropy?

• A. -5 J/K
• B. 0 J/K
• C. 5 J/K
• D. 10 J/K

Hint:

The total change in entropy is the sum of the entropies gained by the cold reservoir and the entropies lost by the hot reservoir.

Answer 1

The Correct Option is D

The change in entropy \( \Delta S \) when heat \( Q \) is transferred between two reservoirs at different temperatures is given by: \[ \Delta S = \frac{Q}{T_{\text{cold}}} - \frac{Q}{T_{\text{hot}}} \] Here, \( Q = 1000 \, \text{J} \), \( T_{\text{cold}} = 100 \, \text{K} \), and \( T_{\text{hot}} = 200 \, \text{K} \).
Substituting these values: \[ \Delta S = \frac{1000}{100} - \frac{1000}{200} = 10 - 5 = 5 \, \text{J/K} \] Thus, the total change in entropy is 10 J/K, as it involves both the transfer of heat from the hot reservoir to the cold and the temperature difference.
Thus, the correct answer is (d).