Question

A Carnot engine whose sink is at $300K$ has an efficiency of $40\%$. By how much should the temperature of the source be increased so as to increase its efficiency by $50\%$ of original efficiency?

• A. $380K$
• B. $275K$
• C. $215K$
• D. $250K$

Answer 1

The correct option is D

$250K$

Step 1: Given

The temperature of the sink, $T_L=300K$

Original efficiency, ${\eta }_1=40\%=0.4$

New efficiency,

$$\begin{gathered} {\eta }_2=40\%+(50\%of40\%) \\ =0.4+50\%(0.4) \\ =0.4+0.2 \\ =0.6 \end{gathered}$$

Step 2: Formula used

Let the initial temperature of the source be $T_H$, then the formula for efficacy is ${\eta }_1=1-\frac{T_L}{T_H}$.

Step 3: Determine the initial temperature of the source

Using the formula for efficiency,

$$\begin{gathered} {\eta }_1=1-\frac{T_L}{T_H} \\ 0.4=1-\frac{300}{T_H} \\ {T}_H=\frac{300}{(1-0.4)} \\ {T}_H=500K \end{gathered}$$

Step 4: Determine the temperature of the source using the new efficiency

Let the new temperature of the source be $T{\text{'}}_H$

$$\begin{gathered} {\eta }_2=1-\frac{T_L}{T{\text{'}}_H} \\ 0.6=1-\frac{300}{T{\text{'}}_H} \\ T{\text{'}}_H=\frac{300}{1-0.6} \\ T{\text{'}}_H=750K \end{gathered}$$

Step 5: Determine the difference in the temperature of the source

$$\begin{gathered} ∆T_H=T{\text{'}}_H-T_H \\ =750-500 \\ =250K \end{gathered}$$

Therefore, the temperature of the source has to be increased by $250K$.

Hence, option D is the correct option.