Question

a carnot engine works between 27 degree Celsius and 527 degree Celsius if it does work of 250 J per cycle then the heat rejected to the sink per cycle

Answer 1

Ans is 400J.by using the formula.      (T2-T1/2) = Q1/Q2

Answer 2

Answer:

The heat rejected to the sink per cycle is equal to 480J.

Explanation:

Given: For a Carnot engine,

The temperature of source, $T_{H}=527^{o}C=273+527=800K$

The temperature of sink, $T_{c}=27^{o}C=273+27=300K$

Work done per cycle $W=250J$

Efficiency of Carnot's engine is given by:

$\eta =1-\frac{T_{c}}{T_{H}}=\frac{W}{Q_{H}}$                                     .....................(1)

where W is work done per cycle

           $Q_{H}$ is heat supplied

Put the values in eq.(1):

$1-\frac{300}{800}=\frac{800}{Q_{H}}$

$Q_{H}(\frac{5}{8} )=800$

$Q_{H}=1280J$

The work done, $W=Q_{H}-Q_{c}$

$Q_{c}=Q_{H}-W\\Q_{c}=1280-800\\Q_{c}=480J$

Therefore, the heat rejected to the sink equals to 480J.