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We have given that heat withdrawn form the source = 10 KJ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
[tex]\eta =\frac{work\ done}{heat\ withdrawn}=\frac{3}{10}=0.3[/tex]
Higher temperature is given by [tex]T_1=273K[/tex]
We have to find the lower temperature [tex]T_2[/tex]
We know that efficiency is also given by
[tex]1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1}[/tex]
So [tex]\frac{273-T_2}{273}=0.3[/tex]
[tex]T_2=191.1K[/tex]
So temperature of the sink will be 191.1 K
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Given that heat withdrawn from the source [tex]Q_{H}[/tex]= 10 KJ
Work done W= 3 KJ
The hot source temperature [tex]T_{1}[/tex] = 273 K
The efficiency of a steam engine with a hot and cold reservoir can be obtained with equations 1 and 2.
η = W/[tex]Q_{H}[/tex] ...................1
η = [tex]\frac{T_{1}-T_{2} }{T_{1} }[/tex]......................2
since the two equations represent the efficiency of a steam turbine, equations 1 and 2 are equal.
Eqn 1 = Eqn 2
W/[tex]Q_{H}[/tex] = [tex]\frac{T_{1}-T_{2} }{T_{1} }[/tex].
Substituting the values in equation above to get [tex]T_{2}[/tex] ;
3/10 = (273 -[tex]T_{2}[/tex] )/273
2370 - 10[tex]T_{2}[/tex] = 3 x 273
10 [tex]T_{2}[/tex] = 2370 -816
[tex]T_{2}[/tex] = 1911 / 10
[tex]T_{2}[/tex] = 191.1 K
Therefore the temperature of the cold sink is 191.1 K
