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11.Thermodynamics
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An ideal gas heat engine operates in Carnot's cycle between $227\,^oC$ and $127\,^oC$ . It absorbs $6.0 \times 10^4\,cal$ at higher temperature. The amount of heat converted into work is equal to
A
$4.8 \times 10^4\,cal$
B
$3.5 \times 10^4\,cal$
C
$1.6 \times 10^4\,cal$
D
$1.2 \times 10^4\,cal$
Solution
$\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=\frac{\mathrm{W}}{\mathrm{Q}}$
$1-\frac{400}{500}=\frac{W}{6 \times 10^{4}}$
$\mathrm{W}=1.2 \times 10^{4} \mathrm{cal}$
Standard 11
Physics
Similar Questions
A student records $\Delta Q,\Delta U$ and $\Delta W$ for a thermodynamic cycle $A \to B \to C \to A.$ Certain entries are missing. Find correct entry in following options
$AB$ | $BC$ | $CA$ | |
$\Delta W$ | $40\,J$ | $30\,J$ | |
$\Delta U$ | $50\,J$ | ||
$\Delta Q$ | $150\,J$ | $10\,J$ |
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