Question

A Carnot engine has the same efficiency between 800 K and 500 K, and \( x>600 \) K and 600 K. The value of \( x \) is:

• A. \( 1000 \text{ K} \)
• B. \( 960 \text{ K} \)
• C. \( 846 \text{ K} \)
• D. \( 754 \text{ K} \)

Hint:

For Carnot engine efficiency problems, always apply the formula \( \eta = 1 - \frac{T_C}{T_H} \) and equate the efficiencies when given two conditions. Solve for the unknown temperature algebraically.

Answer 1

The Correct Option is B

The efficiency of a Carnot engine is given by the formula:

\[\eta = 1 - \frac{T_c}{T_h}\]

where \( \eta \) is the efficiency, \( T_h \) is the temperature of the hot reservoir, and \( T_c \) is the temperature of the cold reservoir. Given:

1. \( T_h = 800 \text{ K}, T_c = 500 \text{ K} \)

2. \( T_h = x \text{ K}, T_c = 600 \text{ K} \)

Both have the same efficiency, so:

\[1 - \frac{500}{800} = 1 - \frac{600}{x}\]

Simplify:

\[\frac{300}{800} = \frac{x-600}{x}\]

\[0.375 = 1 - \frac{600}{x}\]

\[0.375x = x - 600\]

\[0.625x = 600\]

\[x = \frac{600}{0.625}\]

\[x = 960\]

Thus, the value of \( x \) is \( 960 \text{ K} \).

Answer 2

Step 1: Understanding the Efficiency of a Carnot Engine The efficiency \( \eta \) of a Carnot engine is given by the formula: \[ \eta = 1 - \frac{T_C}{T_H} \] where \( T_C \) is the sink temperature (lower temperature) and \( T_H \) is the source temperature (higher temperature).  
Step 2: Setting Up the Efficiency Equations Given that the efficiency remains the same for two different temperature ranges: For the first case: \[ \eta_1 = 1 - \frac{500}{800} \] \[ \eta_1 = 1 - 0.625 = 0.375 \] For the second case, where \( x \) is the unknown higher temperature and the sink temperature is 600 K: \[ \eta_2 = 1 - \frac{600}{x} \] Since the efficiencies are equal: \[ 0.375 = 1 - \frac{600}{x} \] 
Step 3: Solving for \( x \) Rearranging the equation: \[ \frac{600}{x} = 1 - 0.375 \] \[ \frac{600}{x} = 0.625 \] \[ x = \frac{600}{0.625} \] \[ x = 960 \text{ K} \] Thus, the correct value of \( x \) is 960 K.