Question 980: Thermodynamics & KTG

Question

A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency ?

Accepted Answer

We know that efficiency of Carnot Engine
$$ = {{{T_1} - {T_2}} \over {{T_1}}}$$
where, T
1
is temp. of source & T
2
is temp. of sink
$$ 	herefore 0.40 = {{{T_1} - 300} \over {{T_1}}} \Rightarrow {T_1} - 300 = 0.40{T_1}$$
$$0.6{T_1} = 300 \Rightarrow {T_1} = {{300} \over {0.6}} = {{3000} \over 6} = 500K$$
Now efficiency to be increased by 50%
$$ 	herefore $$ $$0.60 = {{{T_1} - 300} \over {{T_1}}} \Rightarrow {T_1} - 300 = 0.6{T_1}$$
$$0.4{T_1} = 300 \Rightarrow {T_1} = {{300} \over {0.4}} = {{300 	imes 10} \over 4} = 750$$
Increase in temp = 750 Ã¢ÂÂ 500 = 250 K

Answer

380 K

Answer

275 K

Answer

325 K

Answer

250  K

Thermodynamics & KTG

A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency ?

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