A610 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?

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

(75°f − 32) × 5/9 = 23.889°c

true tan speed = ang vel x radius