Guest Question.

500 kJ of heat is removed from a constant temperature heat reservoir maintained at 835 K. Heat is received by a system at constant temperature of 720 K. Temperature of the surroundings, the lowest available temperature is 280 K. Determine the net loss of available energy as a result of this irreversible heat transfer. [U.P.S.C. 1992]

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Solution

\(\text{Here, }T_{0} =\text{ 280 K, i.e surrounding temperature.}\\\begin{aligned}\text{Availability for heat reservoir}&= T_{0} · ΔS_{reservoir}\\&= 280 × \frac{500}{835}\\&= 167.67 kJ/kg · K\\\text{Availability for system}&= T_{0} · ΔS_{system}\\&= 280 × \frac{500}{720}\\&= 194.44 kJ/kg · K\\\text{Net loss of available energy}&= (167.67 – 194.44)\\&= – 26.77 kJ/kg · K\\\text{Loss of available energy}&= \textbf{26.77 kJ/kg . K}\end{aligned}\)