Fundamentals of thermodynamics

Question

A certain engine which operates in a Carnot cycle absorbs .3.347 kJ at 400°C, how much work is done by the engine per cycle? The temperature of sink is 100°C.

Difficult

Solution

$\begin{array}{l}\eta =\frac{{T}_{2}-{T}_{1}}{{T}_{2}}=1-\frac{{T}_{1}}{{T}_{2}}=1-\frac{373}{673}=0.446\\ \eta =\frac{\mathrm{\Delta}w}{{q}_{1}}\text{or}\phantom{\rule{1em}{0ex}}\mathrm{\Delta}W=\eta \times {q}_{1}=0.446\times 3.347=1.49\mathrm{kJ}\end{array}$

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