Question

The temperature of the sink of a Carnot engine is \(250\;K\). In order to increase the efficiency of the Carnot engine from \(25\%\) to \(50\%\), the temperature of the source should be increased by

• A. \(\dfrac{1}{3}\times10^3\;K\)
• B. \(\dfrac{1}{2}\times10^3\;K\)
• C. \(200\;K\)
• D. \(\dfrac{1}{6}\times10^3\;K\)

Hint:

For a Carnot engine, \[ \eta=1-\frac{T_{\text{sink}}}{T_{\text{source}}} \] Efficiency increases when the source temperature increases or sink temperature decreases.

Answer 1

The Correct Option is D

Step 1: Write the Carnot efficiency formula.
For a Carnot engine, \[ \eta = 1 - \frac{T_2}{T_1} \] where $T_1$ is the source temperature and $T_2 = 250\ K$ is the sink temperature.
Step 2: Find the original source temperature.
With $\eta_1 = 25\% = 0.25$, \[ 0.25 = 1 - \frac{250}{T_1} \] \[ \frac{250}{T_1} = 0.75 \] \[ T_1 = \frac{250}{0.75} = \frac{1000}{3}\ K \]
Step 3: Set up for the improved efficiency.
Now $\eta_2 = 50\% = 0.5$ with the same sink and a new source $T_1'$: \[ 0.5 = 1 - \frac{250}{T_1'} \]
Step 4: Solve for the new source temperature.
\[ \frac{250}{T_1'} = 0.5 \] \[ T_1' = \frac{250}{0.5} = 500\ K \]
Step 5: Find the required increase in source temperature.
\[ \Delta T = T_1' - T_1 = 500 - \frac{1000}{3} = \frac{1500 - 1000}{3} = \frac{500}{3}\ K \]
Step 6: Express in the option form.
\[ \Delta T = \frac{500}{3}\ K = \frac{1}{6}\times 10^3\ K \] which matches option (4). \[ \boxed{\dfrac{1}{6}\times 10^3\ K} \]