Question

A Carnot's engine working between 277°C and 27°C produces 100 kJ work. Calcula
he heat rejected from the engine.​

Answer 1

Answer:

let us first convert celcius scale to kelvin scale

277 +. 273 = 550 k

27+ 273 = 300 k

so, initially we find the efficiency ,

ie) EFF = 1. - T( lower)

-------------------

T (higher)

= 1. - 300/ 500

500 - 300

= --------------

500

200

= ------ = 2/5

500

EFF = work done

---------------

heat supplied

so ,

2/5 = 100 kJ

------

Q( supplied)

Q( supplied} = 250 kJ.

finally, Q ( rejected ) = Q( supplied ) - workdone.

= ( 250 - 100) kJ

that is , 150 KJ.

so , heat rejected from the engine = 150 KJ.

hope you will be very much clear reading my solution.

Answer 2

Answer:

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Explanation: