the efficiency of a carnot engine is 30% its efficiency is raised to be 60% by how much must the temperature of the source be increased if the sink is at 27 degree celsius
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efficiency of Carnot engine = (1 - T2/T1 )x 100 ℅
where T1 is the temperature of source and T2 is the temperature of sink .
given,
T2 = 27°C = 300 K
now ,
first time ,
efficiency = 30%
30 = (1 - 300/T1) x 100
0.3 = 1 - 300/T1
0.7 = 300/T1
T1 = 3000/7 K
2nd time efficiency = 60%
60 = (1 - 300/T″1) x 100
T″1 =3000/4
now , temperature of source increased= (3000/4 -3000/7) K
=3000(3/28 ) K
=9000/28 K
where T1 is the temperature of source and T2 is the temperature of sink .
given,
T2 = 27°C = 300 K
now ,
first time ,
efficiency = 30%
30 = (1 - 300/T1) x 100
0.3 = 1 - 300/T1
0.7 = 300/T1
T1 = 3000/7 K
2nd time efficiency = 60%
60 = (1 - 300/T″1) x 100
T″1 =3000/4
now , temperature of source increased= (3000/4 -3000/7) K
=3000(3/28 ) K
=9000/28 K
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