Question

In a certain ideal heat engine, 10.00 kJ of heat is withdrawn from the hot source at 273 K and 3.00 kJ of work is generated. What is the temperature of the cold sink?

Answer 1

The temperature of the cold sink would be 191.1 K

Step-by-step explanation:

The steam engine has a cold and hot reservoir;

Given that heat withdrawn from the source
`Q_(H)`= 10 KJ

Work done W= 3 KJ

The hot source temperature
`T_(1)` = 273 K

The efficiency of a steam engine with a hot and cold reservoir can be obtained with equations 1 and 2.

η = W/
`Q_(H)` ...................1

η =
`(T_(1)-T_(2) )/(T_(1) )`......................2

since the two equations represent the efficiency of a steam turbine, equations 1 and 2 are equal.

Eqn 1 = Eqn 2

W/
`Q_(H)` =
`(T_(1)-T_(2) )/(T_(1) )`.

Substituting the values in equation above to get
`T_(2)` ;

3/10 = (273 -
`T_(2)` )/273

2370 - 10
`T_(2)` = 3 x 273

10
`T_(2)` = 2370 -816

`T_(2)` = 1911 / 10

`T_(2)` = 191.1 K

Therefore the temperature of the cold sink is 191.1 K

Answer 2

Temperature of the sink will be 191.1 K

Step-by-step explanation:

We have given that heat withdrawn form the source = 10 KJ

Work done = 3 KJ

We know that efficiency is given by

`\eta =(work\ done)/(heat\ withdrawn)=(3)/(10)=0.3`

Higher temperature is given by
`T_1=273K`

We have to find the lower temperature
`T_2`

We know that efficiency is also given by

`1-(T_2)/(T_1)=(T_1-T_2)/(T_1)`

So
`(273-T_2)/(273)=0.3`

`T_2=191.1K`

So temperature of the sink will be 191.1 K