A heat engine operates with the cold reservoir at temperature $324 K$ . The minimum temperature of the hot reservoir, if the heat engine takes $300 J$ heat from the hot reservoir and delivers $180 J$ heat to the cold reservoir per cycle, is $K$ .

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Solution

From carnot efficiency:
$(1 - \frac{T_{L}}{T_{H}}) = \frac{Q_{1} - Q_{2}}{Q_{1}}$
(NOTE: symbols have their usual meanings)
$(1 - \frac{324}{T_{H}}) = \frac{300 - 180}{300}$
$1 - \frac{2}{5} = \frac{324}{T_{H}}$
$T_{H} = \frac{324 \times 5}{3} = 540 K$

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A heat engine operates with the cold reservoir at temperature 324 K. The minimum temperature of the hot reservoir, if the heat engine takes 300 J heat from the hot reservoir and delivers 180 J heat to the cold reservoir per cycle, is K.

From carnot efficiency: (1−TLTH)=Q1−Q2Q1 (NOTE: symbols have their usual meanings) (1−324TH)=300−180300 1−25=324TH TH=324×53=540 K

A heat engine operates with the cold reservoir at temperature $324 K$ . The minimum temperature of the hot reservoir, if the heat engine takes $300 J$ heat from the hot reservoir and delivers $180 J$ heat to the cold reservoir per cycle, is $K$ .

From carnot efficiency:
$(1 - \frac{T_{L}}{T_{H}}) = \frac{Q_{1} - Q_{2}}{Q_{1}}$
(NOTE: symbols have their usual meanings)
$(1 - \frac{324}{T_{H}}) = \frac{300 - 180}{300}$
$1 - \frac{2}{5} = \frac{324}{T_{H}}$
$T_{H} = \frac{324 \times 5}{3} = 540 K$