Answer:
[tex]T_C=256.2^{\circ}C[/tex]
Explanation:
Given that,
Efficiency of heat engine, [tex]\eta=40\%=0.4[/tex]
Temperature of hot source, [tex]T_H=427^{\circ}C[/tex]
We need to find the temperature of cold sink i.e. [tex]T_C[/tex]. The efficiency of heat engine is given by :
[tex]\eta=1-\dfrac{T_C}{T_H}[/tex]
[tex]T_C=(1-\eta)T_H[/tex]
[tex]T_C=(1-0.4)\times 427[/tex]
[tex]T_C=256.2^{\circ}C[/tex]
So, the temperature of the cold sink is 256.2°C. Hence, this is the required solution.
