An engine operating between temperatures T₁ and T₂ has efficiency 1 5 . When T ₂ is lowered by 45 K , its efficiency becomes 1 2 . Temperatures T ₁ and T ₂ are respectively

Step-by-step solution

The Carnot efficiency formula gives = 1 - T₂/T₁ where T₁ is the source temperature and T₂ is the sink temperature. For initial efficiency = 1/5 , solving 1/5 = 1 - T₂/T₁ yields T₂/T₁ = 4/5 , so T₁ = (5/4)T₂ . When T₂ decreases by 45 K with constant T₁ , the new efficiency is 1/2 = 1 - (T₂ - 45)/T₁ . This gives (T₂ - 45)/T₁ = 1/2 , so T₁ = 2(T₂ - 45) . Equating both expressions: (5/4)T₂ = 2(T₂ - 45) . Multiplying through by 4: 5T₂ = 8T₂ - 360 . Solving yields -3T₂ = -360 , so T₂ = 120 K. Substituting back: T₁ = (5/4