Question

a 100g ice cube at 0 degrees celsius is placed in 650 grams of water at 25 degrees celsius. When the mixture reaches equillibrium, the temperature of the mixture is 11 degrees celsius. from this information find the latent heat of fusion of water ( specific heat of water is 4186 J/kg degree celsius)

Answer 1

Answer:

The latent heat of fusion of water is 334.88 Joules per gram of water.

Explanation:

Let the latent heat of ice be 'x' J/g

1) Thus heat absorbed by 100 gram of ice to get converted into water equals

$Q_1=100\times x$

2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals

$Q_2=100\times 4.186\times 11=4604.6Joules$

Thus total energy needed equals $Q_1+Q_2=100x+4604.6$

3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is

$Q_3=650\times 4.186\times (25-11)\\\\Q_{3}=38092.6Joules$

Now by conservation of energy we have

$Q_1+Q_2=Q_3\\\\100x+4604.6=38092.6\\\\\therefore x=\frac{38092.6-4604.6}{100}=334.88J/g$