Question

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.)

Answer 1

Answer:

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