Q133 / 180 · KCET 2016 · Physics
A Carnot engine working between \( 300 \mathrm{~K} \) and \( 400 \mathrm{~K} \) has \( 800 \mathrm{~J} \) of useful work. The amount of
heat energy supplied to the engine from the source is
heat energy supplied to the engine from the source is
- A\( 2400 \mathrm{~J} \)
- Correct answerB\( 3200 \mathrm{~J} \)
- C\( 1200 \mathrm{~J} \)
- D\( 3600 \mathrm{~J} \)
Solution
Given,
\( T_{1}=400 K ; T_{2}=300 K ; Q_{2}=800 \)
We know \( \eta=1-\frac{T_{2}}{T_{1}} \)
Also \( 1-\frac{T_{2}}{T_{1}}=\frac{Q_{2}}{Q_{1}} \)
\( \Rightarrow 1-\frac{300}{400}=\frac{800}{Q_{1}} \)
\( \Rightarrow \frac{100}{400}=\frac{800}{Q_{1}} \)
\( \Rightarrow Q_{1}=\frac{800 \times 400}{100}=3200 \mathrm{~J} \)
Therefore, amount of heat energy supplied to the engine from source is \( 3200 \mathrm{~J} \).
\( T_{1}=400 K ; T_{2}=300 K ; Q_{2}=800 \)
We know \( \eta=1-\frac{T_{2}}{T_{1}} \)
Also \( 1-\frac{T_{2}}{T_{1}}=\frac{Q_{2}}{Q_{1}} \)
\( \Rightarrow 1-\frac{300}{400}=\frac{800}{Q_{1}} \)
\( \Rightarrow \frac{100}{400}=\frac{800}{Q_{1}} \)
\( \Rightarrow Q_{1}=\frac{800 \times 400}{100}=3200 \mathrm{~J} \)
Therefore, amount of heat energy supplied to the engine from source is \( 3200 \mathrm{~J} \).
Hence, the correct answer is option: 'B'.