Fusion of ice

1. Jul 9, 2010

The legend

1. The problem statement, all variables and given/known data

A piece of ice at-10 centigrade is heated to -1 C using a certain quantity of energy. Then another 20 times as much energy is necessary to finally obtain water. Using that the specific heat of ice is half of the specific heat of 4.2 kJ/(kg oC) of water, determine the heat of fusion of ice from the above measurement data.

2. Relevant equations
Q=mct
Q=mL

3. The attempt at a solution
Assuming heat to be Q
and mass m

Q=mct
Q=m x 2.1 x 9

and also

20Q = m x 2.1 x 1
Equating
m x 2.1 x 9 x 20 = m x 2.1 x 1

(This is wrong since Fusion of ice didnt come into picture... so any help appreciated)
(Also this is not a homework question... just one which I saw in some book )

2. Jul 9, 2010

Tusike

The problem says 20 much more energy is required to get WATER. So your second equation:

20Q = m*2.1*1 is wrong, it's supposed to be:

20Q = m*2.1*1 + mL

-Tusike

3. Jul 9, 2010

The legend

Thanks a lot Tusike .