A Carnot heat engine receives heat at 900 K and rejects the waste heat to the environment at 300 K. The entire work output of the heat engine is used to drive a Carnot refrigerator that removes heat from the cooled space at –15°C at a rate of 295 kJ/min and rejects it to the same environment at 300 K. Determine the rate of heat supplied to the heat engine. (Round the final answer to one decimal place. You must provide an answer before moving to the next part.).The rate of heat supplied to the engine is ___ kJ/min.

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jbiain jbiain · Answer 1 · 2019-09-18T15:31:43+02:00

Answer:

The rate of heat supplied to the engine is 71.7 kJ/min

Explanation:

Data

Engine hot temperature, [tex] T_H [/tex] = 900 K

Engine cold temperature, [tex] T_C [/tex] = 300 K

Refrigerator cold temperature, [tex] T'_C [/tex] = -15 C + 273 = 258 K

Refrigerator hot temperature, [tex] T'_H [/tex] = 300 K

Heat removed by refrigerator, [tex] Q'_{in} [/tex] = 295 kJ/min

Rate of heat supplied to the heat engine, [tex] Q_{in} [/tex] = ? kJ/min

See figure

From Carnot refrigerator coefficient of performance definition

[tex] COP_{ref} = \frac{T'_C}{T'_H - T'_C} [/tex]

[tex] COP_{ref} = \frac{258}{300 - 258} [/tex]

[tex] COP_{ref} = 6.14 [/tex]

Refrigerator coefficient of performance is defined as

[tex] COP_{ref} = \frac{Q'_{in}}{W} [/tex]

[tex] W = \frac{Q'_{in}}{COP_{ref}} [/tex]

[tex] W = \frac{295 kJ/min}{6.14} [/tex]

[tex] W = 48.04 kJ/min [/tex]

Carnot engine efficiency is expressed as

[tex] \eta = 1 - \frac{T_C}{T_H}[/tex]

[tex] \eta = 1 - \frac{300 K}{900 K}[/tex]

[tex] \eta = 0.67[/tex]

Engine efficiency is defined as

[tex] \eta = \frac{W}{Q_{in}} [/tex]

[tex] Q_{in} = \frac{W}{\eta} [/tex]

[tex] Q_{in} = \frac{48.04 kJ/min}{0.67} [/tex]

[tex] Q_{in} = 71.7 kJ/min [/tex]