Question

The Kelvin temperature of the hot reservoir of an engine is twice that of the cold reservoir, and work done by the engine per cycle is 50 J.
Calculate:
(a) the efficiency of the engine,
(b) the heat absorbed per cycle, and
(c) the heat rejected per cycle.

Answer 1

Answer:

a) 50%

b) 100 J

c) 50 J

Explanation:

The cold temperature of the reservoir = $T_{c}$

according to the problem, it is stated that the hot reservoir of an engine is twice that of the cold reservoir, therefore,

the hot temperature of the reservoir $T_{h}$ = $2T_{c}$

The work done by the engine = 50 J

a) The max efficiency obtainable from a heat engine η =  $1 - \frac{T_{c} }{T_{h} }$

since $T_{h}$ = $2T_{c}$, the equation becomes

η =  $1 - \frac{T_{c} }{2T_{c} }$ =

η =  $1 - \frac{1 }{2 }$ = 0.5 = 50%

b) The heat absorbed per cycle will be gotten from

η =  $\frac{W}{Q}$

η is the efficiency of the system = 0.5

where W is the work done = 50 J

Q is the heat absorbed = ?

substituting, we have

0.5 =  $\frac{50}{Q}$

Q = 50/0.5 = 100 J

c) The heat rejected per cycle = 50% of the absorbed heat

==> 0.5 x 100 J = 50 J