Question 987: Thermodynamics & KTG

Question

The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be

Accepted Answer

Efficiency ($$\eta$$) of a carnot engine is given by
 $$\eta  = 1 - {{{T_2}} \over {{T_1}}}$$, where T
1
is the temperature of the source and T
2
is the temperature of the sink.
Here, T
2
= 500 K
$$ 	herefore $$ $$0.5 = 1 - {{500} \over {{T_1}}} \Rightarrow {T_1} = 1000K$$
Now, $$\eta  = 0.6 = 1 - {{{T_2}'} \over {1000}}$$ (T
2
' is the new sink
temperature)
$$ \Rightarrow {T_2}' = 400\,K$$

Answer

100 K

Answer

600 K

Answer

400 K

Answer

500 K

Thermodynamics & KTG

The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be

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