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
[tex]Q_1=100\times x[/tex]
2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals
[tex]Q_2=100\times 4.186\times 11=4604.6Joules[/tex]
Thus total energy needed equals [tex]Q_1+Q_2=100x+4604.6[/tex]
3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is
[tex]Q_3=650\times 4.186\times (25-11)\\\\Q_{3}=38092.6Joules[/tex]
Now by conservation of energy we have
[tex]Q_1+Q_2=Q_3\\\\100x+4604.6=38092.6\\\\\therefore x=\frac{38092.6-4604.6}{100}=334.88J/g[/tex]
