A Carnot engine has an efficiency of 59% and performs 2.5 × 104 J of work in each cycle. (a) How much heat does the engine extract from its heat source in each cycle. (b) Suppose the engine exhausts heat at room temperature (20.0°C). What is the temperature of its heat source?

Solution 15E W The efficiency of a Carnot’s engine is given by = Q a…..(1), where Q isathe amount of heat supplied to a system and W is the work done by the system. Given, = 59% = 0.59 W = 2.5 × 10 J (a) The heat extracted by the engine, Q = W (from equation 1) a 4 Q a 2.5×10 = 4.23 × 10 J 0.59 3 Q a 42.3 × 10 J Q a 42.3 kJ The engine extracts a heat of 42.3 kJ from its heat source. (b) Let the temperature of the heat source be H . Room temperature = 20.0 C = 273 + 20K = 293 K 293 Therefore, efficiency = 1 T H 0.59 = 1 293 TH T H = 293K 0.41 T = 714 K H 0 T H = 714 273 C T H = 441 C0 0 Therefore, the temperature of the heat source will be 441 C .