A perfect Carnot engine utilizes an ideal gas. The source tempera… - Vidyadip

Question

A perfect Carnot engine utilizes an ideal gas. The source temperature is $500K$ and sink temperature is $375K$. If the engine takes $600K$ cal per cycle from the source, compute:

The efficiency of the engine.
Work done per cycle.
Heat rejected to the sink per cycle.

✓

Answer

Here $T_1 = 500K$ $T_2 = 375K Q_1 =$ Heat absorbed per cycle $= 600\ kcal$

$\therefore$ Using the relation,
$\eta=1-\frac{\text{T}_2}{\text{T}_1},$ we get
$\eta=\frac{\text{T}_1-\text{T}_2}{\text{T}_1}$
$=\frac{500-375}{500}$
$=\frac{125}{500}=0.25$
$\eta\%=0.25\times100$
$=25\%$

Let $W =$ work done per cycle

$\therefore$ Using the relation
$\eta=\frac{\text{W}}{\text{Q}_1},$ we get
$\text{W}=\eta\text{Q}_1=0.25\times600\text{ kcal}=150\text{ kcal}$
$=150\times10^3\times4.2\text{J}$
$=6.3\times10^5\text{J}$

Let $Q_2 =$ Heat rejected to the sink

$\therefore$ Using the relation
$\text{W}=\text{Q}_1-\text{Q}_2,$ we get
$\text{Q}_2=\text{Q}_1-\text{W}$
$=600-150=450\text{ kcal}$

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