Answer:
960 K
Explanation:
The efficiency (η) of a Carnot engine is equal to one minus the ratio of the cold temperature (Tc) to the hot temperature (Th).
[tex]\Large \text{$ \eta=1- $}\huge \text{$ \frac{T_C}{T_H} $}[/tex]
Initially, the high temperature is 800 K and the efficiency is 40%. Therefore, the cold temperature is:
[tex]\Large \text{$ 0.40=1- $}\huge \text{$ \frac{T_C}{800\ K} $}\\\\\Large \text{$ 0.60= $}\huge \text{$ \frac{T_C}{800\ K} $}\\\\\Large \text{$ T_C=480\ K $}[/tex]
To increase the efficiency to 50%, the high temperature must become:
[tex]\Large \text{$ 0.50=1- $}\huge \text{$ \frac{480\ K}{T_H} $}\\\\\Large \text{$ 0.50=\ $}\huge \text{$ \frac{480\ K}{T_H} $}\\\\\Large \text{$ T_H=960\ K $}[/tex]
