For an ideal heat engine, the temperature of the source is 127°C. In order to have 60% efficiency the temperature of the sink should be ________°C. (Round off to the nearest integer)

For an ideal heat engine (Carnot engine), the efficiency is given by $$\eta = 1 - \frac{T_{\text{sink}}}{T_{\text{source}}}$$, where temperatures are in Kelvin.

The source temperature is $$T_{\text{source}} = 127°\text{C} = 127 + 273 = 400$$ K. The required efficiency is $$\eta = 60\% = 0.6$$.

Substituting: $$0.6 = 1 - \frac{T_{\text{sink}}}{400}$$, which gives $$\frac{T_{\text{sink}}}{400} = 0.4$$, so $$T_{\text{sink}} = 160$$ K.

Converting back to Celsius: $$T_{\text{sink}} = 160 - 273 = -113$$°C.

The temperature of the sink is $$\boxed{-113}$$°C.