Question

A Carnot engine whose low temperature reservoir is at 350 K has an efficiency of 50%. It is desired to increase this to 60%. If the temperature of the low temperature reservoir remains constant, then the temperature of high temperature reservoir must be increased by how many degrees?

• A. 15
• B. 175
• C. 100
• D. 50
• E. 120

Hint:

To increase the efficiency of a Carnot engine, you can either increase the source temperature or decrease the sink temperature.

Answer 1

The Correct Option is B

Concept:
The efficiency of a Carnot engine is $\eta = 1 - \frac{T_L}{T_H}$, where $T_L$ is the sink temperature and $T_H$ is the source temperature in Kelvin.

Step 1: Find the initial source temperature ($T_{H1}$).
Given $\eta_1 = 0.5$ and $T_L = 350$ K. \[ 0.5 = 1 - \frac{350}{T_{H1}} \implies \frac{350}{T_{H1}} = 0.5 \implies T_{H1} = 700 \text{ K} \]

Step 2: Find the new source temperature ($T_{H2}$).
Given $\eta_2 = 0.6$ and $T_L = 350$ K. \[ 0.6 = 1 - \frac{350}{T_{H2}} \implies \frac{350}{T_{H2}} = 0.4 \implies T_{H2} = \frac{350}{0.4} = 875 \text{ K} \]

Step 3: Calculate the increase.
Increase in temperature $= T_{H2} - T_{H1} = 875 - 700 = 175$ degrees.