Question

an engine has been designed to work between source and sink at temperatures 177 c and 27 c respectively. if energy input is 3600j. what is the work done by the engine?

Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

An engine has been designed to work between source and sink at temperatures 177 c and 27 c respectively. If energy input is 3600 J. What is the work done by the engine?

$\huge\underline{\underline{\bf \orange{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• ${\sf Sink\: Temperature (T_1)=177+273\implies 450K}$
• ${\sf Source\: Temperature (T_2)=27+273\implies 300K}$
• ${\sf Energy\:input(Q_1)=3600J}$

$\large\underline{\underline{\sf To\:Find:}}$

• Work Done by engine

✪$\large{\boxed{\bf \blue{\dfrac{Q_1}{Q_2}=\dfrac{T_1}{T_2}} }}$

$\implies{\sf \dfrac{3600}{Q_2}=\dfrac{450}{300}}$

$\implies{\sf Q_2=\dfrac{3600×300}{450}}$

$\implies{\sf Q_2=\dfrac{1080000}{450} }$

$\implies{\bf\red{Work\:( Q_2)=2400\:J} }$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

Work done by the engine is ${\bf \red{2400\:J}}$ .

Answer 2

W=1200 J

Explanation:

Given:

• temperature of the hotter reservoir, $T_H=177+273=450\ K$

• temperature of the colder reservoir, $T_L=27+273=300\ K$

• energy input to the heat engine, $Q_{in}=3600\ J$

Now the Carnot efficiency of the engine:

$\eta_c=1-\frac{T_L}{T_H}$

$\eta_c=1-\frac{300}{450}$

$\eta_c=33.33\%$

So the work done:

$W=Q_{in}.\eta_c$

$W=3600\times 0.33$

$W=1200\ J$

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TOPIC: heat engine

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