Question

a refrigerator takes heat from water at 0°c and transfer it to room at 27°c. if 100 kg of water is coverted in ice at 0°c then calculate the work done. (latent heat of ice is
$3.4 \times {10}^{5} j \: kg$
)​

Answer 1

$\huge{\underline{\underline{\mathbf{Answer:-}}}}$

$3.36 \times {10}^{6} j$

________________________________

$\huge{\underline{\underline{\mathbf{Explanation:-}}}}$

Coefficient of performance (COP) =

$\large\boxed{ \frac{ t_{2} }{ t_{1} - t_{2} } }$

✍$\large\bold{ \frac{273}{300 - 273} }$

✍$\large\bold{ \frac{273}{27} }$

and,

$\large\boxed{w = \frac{ q_{2} }{cop} = \frac{ml}{cop} }$

✍$\large\bold{ \frac{100 \times 3.4 \times {10}^{5} }{ \frac{273}{27} } }$

✍$\large\bold{ \frac{100 \times 3.4 \times {10}^{5} \times 27 }{273} }$

$\large\boxed{3.36 \times {10}^{6} j}$

_______________________________

Answer 2

Coefficient of performance(cop)=t2/t1-t2

=273/300-273= 273/27

and

w= q2/ cop= ml/cop

= 100×3.4×100000

273

27

= 100×3.4×100000×27

273

= 3.36×1000000 J