CBSE Class 11-science Answered

Maximum efficiency of an engine working between temperatures T₂ and T₁ is given by the fraction of the heat absorbed by an engine which can be converted into work is known as efficiency of the heat engine.

Mathematically,
Efficiency, ? = (T₂ – T₁) / T₂ = 1/5
5 T₂ -5 T₁ = T₂
Therefore, T₂ = 1.25 T₁ ……………………….. (1)
Where T₂ is the source temperature and T₁ is the sink temperature.

If the sink temperature (T₁) is reduced by 70^o C, the efficiency is doubled
?= T₂ – (T₁ – 70) / T₂ = 2x1/5 = 2/5
5 T₂ – 5 T₁ + 350 = 2 T₂
5 T₂ – 2 T₂ – 5 T₁ + 350 = 0
3T₂ – 5 T₁ + 350 = 0
Substituting the value of T₂ from equation (1), we will get,
3(1.25 T₁) – 5 T₁ + 350 = 0
3.75T₁ – 5 T₁ + 350 = 0
350 = 1.25 T₁
T₁ = 300 K
Therefore, T₂ = 1.25 X 300 = 375 K

Hence, temperature of the source and the sink is 375 K and 300 K respectively.