Question

A perfect Carnot engine utilizes an ideal gas. The source temperature is 500 and sink temperature is 375K. If the engine takes 600K cal per cycle from the source, compute :
a) the efficiency of the engine
b) work done per cycle
c) heat rejected to the sink per cycle​

Answer 1

$\Huge {\orange {\bf {\underline {\underline {ÀnSwer}}}:-}}$

a) the efficiency of the engine is 25%

b) work done per cycle is 6.3 × 10⁵J

c) heat rejected to the sink per cycle is 450kcal

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Explaination :-

⟶ Gíven :-

• $\sf T_1$= 500k.
• $\sf T_2$= 375k.
• Heat absiorbed per cycle, $\sf \varrho_{1}$= 600K cal.

⟶ To fínd :-

a) the efficiency of the engine

b) work done per cycle

c) heat rejected to the sink per cycle

⟶ Solutíon :-

a) From the relation,

$\longrightarrow \sf \eta = 1 - \frac{ T_{1} }{T_{2}} \\ \\ \longrightarrow \sf \eta = \frac{ T_{1} - T_{2}}{T_{2}} \\ \\ \longrightarrow \sf \eta = \frac{500 - 375}{500} = \frac{125}{500} \\ \\ \longrightarrow \sf \eta = 0.25 \\ \\ \longrightarrow \sf \eta \% = 0.25 \times 100 = 25\%$

$\therefore$ the efficiency of the engine is 25% .

b) suppose W = work done per cycle

using the relation,

.i.e.,

$\longrightarrow \sf \eta = \frac{W}{ \varrho _{1} } \\ \\ \longrightarrow \sf W = \eta \varrho _{1} \\ \\ \longrightarrow \sf \: W = 0.25 \times 600 \: kcal \\ \\ \longrightarrow \sf \: W = 150 \: kcal \\ \\ \longrightarrow \sf W = 150 \times {10}^{3} \times 4.2j \\ \\ \longrightarrow \sf W = 6.3 \times {10}^{5} j$

$\therefore$ Work done per cycle is 6.3 × 10⁵J.

c) Suppose $\sf \varrho_{2}$= heat rejected to the sink

therefore from the relation

.i.e.,

$\longrightarrow \sf W = \varrho_{1} - \varrho_{2} \\ \\ \longrightarrow \sf \varrho_{2} = \varrho_{1} - W \\ \\ \longrightarrow \sf \varrho_{2} = 600 - 150 \\ \\ \longrightarrow \sf \varrho_{2} =450kcal$

thus, heat rejected to the sink per cycle is 450 Kcal