4. (15 points) The efficiency of an ideal heat engine is $40 \%$. What will it become if the temperature of the heater is increased by $40 \%$, and the temperature of the cooler is decreased by $40 \%$?
Answer: $\approx 74 \%$.

Solution. The efficiency of an ideal heat engine: $\eta=1-\frac{T_{X}}{T_{H}}$. That is, initially the ratio of the temperatures of the refrigerator and the heater: $\frac{T_{X}}{T_{H}}=1-0.4=0.6$. After the changes:

$$
\eta_{2}=1-\frac{0.6 T_{X}}{1.4 T_{H}}=1-\frac{0.6 \cdot 0.6}{1.4} \approx 0.74 \approx 74 \% .
$$