Question

An engine working between the temperatures TH = 145ºC and TC = 30 ºC works at the maximum efficiency possible. It must do the work of lifting 250kg a height of 165m. Determine the work done, the efficiency of the engine and QH and QC. Remember that QH = QC + W

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Temperature of the hot reservoir = T_h = 145 ^oC = 145 + 273 K = 418 K

Temperature of the cold reservoir = T_c = 30 ^oC = 30 + 273 K = 303 K

Efficiency of the engine =

The engine is working at maximum efficiency which is the carnot efficiency therefore,

= 0.2751

Mass of the weight lifted = M = 250 kg

Height the weight is lifted by = H = 165 m

Work done by the engine = W

The engine does a work of lifting 250 kg to a height of 165 m.

W = MgH

W = (250)(9.81)(165)

W = 4.047 x 10⁵ J

Heat input to the engine = Q_h

Q_h = 1.471 x 10⁶ J

Heat rejected by the engine = Q_c

Q_h = Q_c + W

1.471x10⁶ = Q_c + 4.047x10⁵

Q_c = 1.066 x 10⁶ J

a) Work done by the engine = 4.047 x 10⁵ J

b) Efficiency of the engine = 0.2751 = 27.51%

c) Heat input to the engine = Q_h = 1.471 x 10⁶ J

d) Heat rejected by the engine = Q_c = 1.066 x 10⁶ J