Question

In a Carnot engine, the temperature of the sink is 350 K. If the efficiency of the engine is 50%, the temperature of the source should be:

• A. 700 K
• B. 750 K
• C. 800 K
• D. 900 K
• E. 1000 K

Hint:

The efficiency of a Carnot engine is related to the temperatures of the hot and cold reservoirs. The formula \( \eta = 1 - \frac{T_{{sink}}}{T_{{source}}} \) helps calculate the required temperature of the source given the efficiency and temperature of the sink.

Answer 1

The Correct Option is A

The efficiency \( \eta \) of a Carnot engine is given by the following equation: \[ \eta = 1 - \frac{T_{{sink}}}{T_{{source}}} \] where: - \( \eta \) is the efficiency of the engine,
- \( T_{{sink}} \) is the temperature of the sink (cold reservoir),
- \( T_{{source}} \) is the temperature of the source (hot reservoir).
We are given: - \( \eta = 50% = 0.5 \),
- \( T_{{sink}} = 350 \, {K} \).
Substitute these values into the efficiency equation: \[ 0.5 = 1 - \frac{350}{T_{{source}}} \] Solving for \( T_{{source}} \): \[ \frac{350}{T_{{source}}} = 1 - 0.5 = 0.5 \] \[ T_{{source}} = \frac{350}{0.5} = 700 \, {K} \] Thus, the temperature of the source is \( 700 \, {K} \). \[ \boxed{700 \, {K}} \]