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A 610 W Carnot engine operates between constant-temperature reservoirs at 142°C and 69.9°C. What is the rate at which energy is (a) taken in by the engine as heat and (b) exhausted by the engine as heat?

Respuesta :

Answer:

rate at which energy taken is 3511.19 W

rate at which heat exhausted is 2901.19 W

Explanation:

Given data

power = 610 W

Temperatures T = 142°C  = 142 + 273 = 415 K

Temperatures T2 = 69.9°C  = 69.9 + 273 = 342.9 K

to find out

rate at which energy taken and heat exhausted

solution

we know the equation of efficiency of engine that is = 1 - (T2 / T1)

so efficiency = 1 - (342.9 / 415 )

efficiency is 0.17373

and we know efficiency = energy output / energy input

efficiency = energy output / energy input

0.17373 = 610 / energy input

energy input = 3511.19

so rate at which energy taken is 3511.19 W

and rate at which heat exhausted is 3511.19  - power

rate at which heat exhausted is 3511.19  - 610

rate at which heat exhausted is 2901.19 W

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