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PART - 2 CH - 11 Thermodynamics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 11 Thermodynamics2 Marks
Question
A heat engine takes heat $Q_1$ at a temperature of 500 K and gives heat $Q_2$ at a temperature of 300 K to the sink and does $8 \times 10^5$ joules of work. Calculate the magnitude of $Q_1$ and $Q_2$.
✓

Answer

Given that :
$\begin{aligned}T_1 & =500 K \\T_2 & =300 K \\W & =8 \times 10^5 \text { Joule }\end{aligned}$
(i) Heat absorbed from the heat source :
$\begin{aligned}\eta & =\frac{T_1-T_2}{T_1}=\frac{W}{Q_1} \\Q_1 & =\frac{WT_1}{T_1-T_2} \\Q_1 & =\frac{8 \times 10^5 \times 500}{(500-300)} \\Q_1 & =\frac{8 \times 10^5 \times 500}{200}\end{aligned}$
$=20 \times 10^5$ joule
(ii) Dissipated heat
$ \begin{aligned} Q_2 & =Q_1-W \quad\left[\because Q_1-Q_2=W\right] \\ & =20 \times 10^5-8 \times 10^{\frac{5}{3}} \\ & =12 \times 10^5 \text { Joule }\end{aligned} $

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