First slide

Heat Engine; Carnot Engine and refrigerator

Question

An ideal gas heat engine operates in carnot’s cycle between $227^{o} c$ and $127^{o} c$ . If absorbs ${6×10}^{4} cals$ of heat at higher temperature amount of heat converted to work is

Moderate

A

${4.8×10}^{4} cals$
B

${1.2×10}^{4} cals$
C

${2.4×10}^{4} cals$
D

${6×10}^{4} cals$

Solution

$\begin{array}{rcl} {Temperature T}_{1} =227+273=500 K (converting degree celsius into kelvin) \\ {:T}_{2} =127+273=400 K \\ e f f i e c i e n c y o f h e a t e n g i n e i s \\ η=1- \frac{T_{2}}{T_{1}} ---(1) \\ s u b s t i t u t e g i v e n t e m p e r a t u r e s i n e q u a t i o n (1) \\ \Rightarrow & η = 1 - \frac{400}{500} \\ \Rightarrow & η= \frac{1}{5} ---(2) \\ a l s o e f f i c i e n c y i s η= \frac{Work done (w)}{heat given to engine(Q)} ---(3) \\ s u b s t i t u t e g i v e n Q & = & {6×10}^{4} calorie, and equation (2) in (3) \\ \Rightarrow & \frac{1}{5} = \frac{W}{{6×10}^{4}} \\ W= \frac{{6×10}^{4}}{5} \\ {W=1.2×10}^{4} cal \\ a m o u n t o f h e a t c o n v e r t e d t o w o r k & = & 1.2 {\times10}^{4} cal \end{array}$

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