KCET · Physics · Thermodynamics
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
- A \( 2400 \mathrm{~J} \)
- B \( 3200 \mathrm{~J} \)
- C \( 1200 \mathrm{~J} \)
- D \( 3600 \mathrm{~J} \)
Answer & Solution
Correct Answer
(B) \( 3200 \mathrm{~J} \)
Step-by-step Solution
Detailed explanation
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 \( 3200 \mathrm{~J} \).
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