 Physics
6 August, 22:26

# An ideal heat-engine is to be used in an environment where the ambient temperature is 23.5 °C. What should be the minimum temperature of the hot heat reservoir in order to reach at least 41.2 percent efficiency with the heat-engine? (Give your answer in Celsius.)

+3
1. 6 August, 23:23
0
The minimum temperature of the hot reservoir = 231.25 °C

Explanation:

η₁ = (1 - T (c) / T (h)) * 100 ... equation 1

Where η₁ = efficiency of the heat engine in percentage, T (c) = Temperature at which the low temperature reservoir operates, T (h)) = Temperature at which the high temperature reservoir operates.

Making T (h) the subject of the equation in equation 1

T (h) = T (c) / {1 - (η₁ / 100) }

where η₁ = 41.2%, T (c) = 23.5 °C = 23.5 + 273 = 296.5 K

∴T (h) = 296.5/{1 - (41.2/100) }

T (h) = 296.5 / (1 - 0.412)

T (h) = 296.5/0.588

T (h) = 504.25 K

Convert to temperature in Celsius

T (h) = 504.25 - 273 = 231.25 °C

The minimum temperature of the hot reservoir = 231.25 °C