Question

A Carnot engine of 50 % efficiency takes heat from a source of 400 K. To change efficiency to 70 % without changing the sink temperature, the new temperature of source should be (nearer to)

• A. 765 K
• B. 525 K
• C. 800 K
• D. 667 K

Hint:

In Carnot engine problems, always solve for the unchanged temperature reservoir first!

Answer 1

The Correct Option is D

Step 1: Carnot efficiency formula.
For a Carnot engine $\eta = 1 - \dfrac{T_{2}}{T_{1}}$, where $T_{1}$ is the source and $T_{2}$ is the sink (in kelvin).

Step 2: Use the first case to find the sink.
Given $\eta_{1}=0.50$ and $T_{1}=400$ K: \[ 0.50 = 1 - \frac{T_{2}}{400} \]
Step 3: Solve for the sink temperature.
\[ \frac{T_{2}}{400} = 0.50 \quad\Rightarrow\quad T_{2} = 200 \text{ K} \]
Step 4: Set up the new case.
Now we want $\eta_{2}=0.70$ with the same sink $T_{2}=200$ K. Let the new source be $T_{1}'$. \[ 0.70 = 1 - \frac{200}{T_{1}'} \]
Step 5: Rearrange.
\[ \frac{200}{T_{1}'} = 0.30 \]
Step 6: Solve for the new source.
\[ T_{1}' = \frac{200}{0.30} \approx 667 \text{ K} \]So the new source temperature is about $667$ K, which is option 4.
\[ \boxed{T_{1}' \approx 667 \text{ K}} \]