Question

A heat engine operates at 30% of its maximum possible efficiency and needs to do 995 J of work. Its cold reservoir is at 22 ºC and its hot reservoir is at 610 ºC. (a) How much energy does it need to extract from the hot reservoir? (b) How much energy does it deposit in the cold reservoir?

Answer 1

Answer:

(a) The energy  extracted from the hot reservoir (Qh) is 3316.67J

(b) The energy deposited in the cold reservoir (Qc) is 2321.67J

Explanation:

Part (a) The energy  extracted from the hot reservoir (Qh)

e = W/Qh

where;

e is the maximum efficiency of the system = 30% = 0.3

W is the the work done on the system = 995 J

Qh is the heat absorbed from the hot reservoir

Qh = W/e

Qh = 995/0.3

Qh = 3316.67J

Part (b) The energy deposited in the cold reservoir (Qc)

e = W/Qh

W = Qh - Qc

where;

Qc is the heat deposited in the cold reservoir

e = (Qh - Qc)/Qh

Qh - Qc = e*Qh

Qc = Qh - e*Qh

Qc = 3316.67J - 0.3*3316.67J

Qc = 3316.67J - 995J

Qc = 2321.67J