Question

To increase the efficiency of a Carnot engine working between a source at 500 K and a sink at 300 K by 10%, with the same sink temperature the source temperature should be increased to

• A. 550 K
• B. 600 K
• C. 700 K
• D. 750 K
• E. 800 K

Hint:

Logic Tip: Note the specific phrasing "increased by 10%". In the context of engine efficiencies, this almost always refers to an absolute increase (+0.10) rather than a relative increase ($0.40 \times 1.10 = 0.44$). If it meant the latter, the answer would not be a clean integer.

Answer 1

The Correct Option is B

Concept:
The efficiency ($\eta$) of a Carnot engine depends only on the absolute temperatures of the hot source ($T_H$) and the cold sink ($T_C$): $$\eta = 1 - \frac{T_C}{T_H}$$ An increase in efficiency by 10% means adding 0.10 to the original efficiency value.
Step 1: Calculate the initial efficiency ($\eta_1$).
Given initial temperatures: $T_H = 500\text{ K}$ $T_C = 300\text{ K}$ $$\eta_1 = 1 - \frac{300}{500}$$ $$\eta_1 = 1 - \frac{3}{5} = 1 - 0.6 = 0.4$$ The initial efficiency is 40%.
Step 2: Determine the target new efficiency ($\eta_2$).
The problem states the efficiency is increased by 10% (an absolute percentage point increase). $$\eta_2 = \eta_1 + 0.10$$ $$\eta_2 = 0.40 + 0.10 = 0.50$$ The new efficiency is 50%.
Step 3: Calculate the new source temperature ($T_H'$).
The sink temperature $T_C$ remains constant at 300 K. $$\eta_2 = 1 - \frac{T_C}{T_H'}$$ $$0.50 = 1 - \frac{300}{T_H'}$$ Rearrange to solve for $T_H'$: $$\frac{300}{T_H'} = 1 - 0.50$$ $$\frac{300}{T_H'} = 0.50$$ $$T_H' = \frac{300}{0.50} = 600\text{ K}$$