Maximum efficiency of an engine working between temperatures T2 and T1 is given by the fraction of the heat absorbed by the engine, which can be converted into work. Mathematically, Efficiency, η =
(T2–T1)
T2
= 1/6 6T2−6T1 = T2 Therefore, T2 = 1.2 T1 ………………….. (1) Where T2 is the source temperature and T1 is the sink temperature If the sink temperature (T1) is reduced by 62°C, then the efficiency gets doubled, i.e. η = T2–
(T1–62)
T2
= 2 x 1/6 = 2/6 6T2–6T1+372 = 2T2 6T2–2T2–6T1 + 372 = 0 4T2–6T1 + 372 = 0 Substituting the value of T2 from equation (1), we get 4(1.2T1)–6T1 + 372 = 0 4.8T1–6T1 + 372 = 0 372 = 1.2T1 T1 = 310 K Therefore, T2 = 1.2 x 310 = 372 K Hence, the temperatures of the source and the sink are 372 K and 310 K, respectively. Source temperature = 372 - 273 = 99°C