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.

a) 50%

b) 100 J

c) 50 J

Explanation:

The cold temperature of the reservoir =

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 =

The work done by the engine = 50 J

a) The max efficiency obtainable from a heat engine η =  

since = , the equation becomes

η =   =

η =   = 0.5 = 50%

b) The heat absorbed per cycle will be gotten from

η =  

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 =  

Q = 50/0.5 = 100 J

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

==> 0.5 x 100 J = 50 J

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//