Question

For a certain ideal Carnot engine, the hot reservoir is 35°C higher than the cold reservoir. If this engine is to have an efficiency of 20%, what must be the temperature of the hot reservoir?

Answer 1

Answer:

Temperature of the hot reservoir is 1540K

Explanation:

$E= 1- \frac{T_{c}}[tex]{T_h}=308+{T_c}\\Efficiency of a carnot engine is given by the aboveTc=temperature of the cold reservoirTh= temperature of the hot reservoirK=273+ 35 (convert 35°C to kelvin)K=308k{T_h}={T_c}+308-----------------------(equation 1)20%=1-{T_c}/{T_h}$

0.2=({T_c}+308-{T_c})/{T_c}+308

.2({T_c}+61.6=308

0.2{T_c}=246.4

{T_c}=1232

recall from equation 1

{T_h}=308+1232

{T_h}=1540K