Question

Find the required work to transfer 1200 J of heat from the sink (cold body) at -90° C to
source (hot body) at 35°C in the case of refrigerator.

Answer 1

Given that,

Heat = 1200 J

Temperature T₁ = 35°C

Temperature T₂ = -90°C

We need to calculate the required work

Using formula of work

$\dfrac{Q}{W}=\dfrac{T_{1}}{T_{1}-T_{2}}$

Where, Q = heat

$T_{1}$ = source temperature

$T_{2}$ = sink temperature

Put the value into the formula

$\dfrac{1200}{W}=\dfrac{35}{35+90}$

$W= \dfrac{1200(35+90)}{35}$

$W=4285.7\ J$

Hence, The required work is 4285.7 J.

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Topic : refrigerator

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