Question

A heat engine has a maximum possible efficiency of 0.780. If it operates between a deep lake with a constant temperature of-24.8°C, what is the temperature of the hot reservoir (in Kelvins)?

Answer 1

Answer : The temperature of the hot reservoir (in Kelvins) is 1128.18 K

Explanation :

Efficiency of carnot heat engine : It is the ratio of work done by the system to the system to the amount of heat transferred to the system at the higher temperature.

Formula used for efficiency of the heat engine.

$\eta =1-\frac{T_c}{T_h}$

where,

$\eta$ = efficiency = 0.780

$T_h$ = Temperature of hot reservoir = ?

$T_c$ = Temperature of cold reservoir = $-24.8^oC=273+(-24.8)=248.2K$

Now put all the given values in the above expression, we get:

$\eta =1-\frac{T_c}{T_h}$

$0.780=1-\frac{248.2K}{T_h}$

$T_h=1128.18K$

Therefore, the temperature of the hot reservoir (in Kelvins) is 1128.18 K