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In a certain ideal heat engine, 10.00 kJ of heat is withdrawn from the hot source at 273 K and 3.00 kJ of work is generated.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

Answer:

Temperature of the sink will be 191.1 K

Explanation:

We have given that heat withdrawn form the source = 10 KJ

Work done = 3 KJ

We know that efficiency is given by

$\eta =\frac{work\ done}{heat\ withdrawn}=\frac{3}{10}=0.3$

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

$1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1}$

So $\frac{273-T_2}{273}=0.3$

$T_2=191.1K$

So temperature of the sink will be 191.1 K

Answer 2

Answer:

The temperature of the cold sink would be 191.1 K

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Given  that heat withdrawn from the source $Q_{H}$= 10 KJ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 = $\frac{T_{1}-T_{2} }{T_{1} }$......................2

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

Eqn 1 = Eqn 2

W/$Q_{H}$ =  $\frac{T_{1}-T_{2} }{T_{1} }$.

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