An ideal Carnot engine operates with an initial efficiency of 1 6 . When the temperature of its source is increased by 90^ C , the efficiency of the engine is doubled. The temperature of the sink is:

Step-by-step solution

Let the initial source temperature be T₁ and the sink temperature be T₂ in Kelvin. The initial efficiency is given by: ₁ = 1 - T₂ T₁ = 1 6 T₂ T₁ = 5 6 T₁ = 1.2 T₂ When the source temperature is increased by 90^ C , the change in Kelvin is also 90 K . The new source temperature is T₁ + 90 . The new efficiency is doubled, so ₂ = 2 1 6 = 1 3 . ₂ = 1 - T₂ T₁ + 90 = 1 3 T₂ T₁ + 90 = 2 3 T₁ + 90 = 1.5 T₂ Substituting T₁ = 1.2 T₂ into the equation: 1.2 T₂ + 90 = 1.5 T₂ 0.3 T₂ = 90 T₂ = 300 K Converting the sink temperatur