COMEDK · Physics · 13. Thermodynamics
If a Carnot heat engine has a sink at a temperature of \(27^\circ\text{C}\) and takes \(100\text{ kcal}\) of heat from a source at a temperature of \(627^\circ\text{C}\), the work done (in J) is:
- A \(0.28 \times 10^6\)
- B \(2.8 \times 10^6\)
- C \(28 \times 10^6\)
- D \(0.028 \times 10^6\)
Answer & Solution
Correct Answer
(A) \(0.28 \times 10^6\)
Step-by-step Solution
Detailed explanation
\(T_2 = 27 + 273 = 300\,\text{K}\), \(T_1 = 627 + 273 = 900\,\text{K}\) \(\eta = 1 - \dfrac{T_2}{T_1} = 1 - \dfrac{300}{900} = \dfrac{2}{3}\) \(Q_1 = 100\,\text{kcal} = 10^5 \times 4.2\,\text{J} = 4.2 \times 10^5\,\text{J}\)…
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