Question

A refrigerator converts 100g of water at 20 C to ice at - 10 C in 73.5 mins.
Calc the average rate of heat extraction in watt.
specify heat capacity of water is 4.2 J g^-1 K^-1.specific latent heat capacity of ice is 336 J g ^-1

Answer 1

@Michele_Laino

Answer 2

here we have to compute the amount of heat which has to be extracted, first

Answer 3

How to do so..??

Answer 4

such amount of heat is:
\[\large Q = Lm + cm\Delta \theta = \left( {336 \cdot 100} \right) + \left( {4.2 \cdot 100 \cdot 30} \right) = ...?\]

Answer 5

46200

Answer 6

correct!

Answer 7

so, the requested average rate \(r\), is given by the subsequent computation:
\[\Large r = \frac{Q}{{\Delta t}} = \frac{{46200}}{{\left( {73.5 \cdot 60} \right)}} = ...Watt\]

Answer 8

10.47

Answer 9

since we need to measure time in seconds, here is why I wrote \(73.5 \cdot 60\)

Answer 10

10.47 Watts..?

Answer 11

correct!

Answer 12

73.5*60 = 4410

Answer 13

we can write:
\(r=10.5 \; watts\)