Calculating Heat of Fusion of Ice from Data

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SUMMARY

The discussion focuses on calculating the heat of fusion of ice based on a problem involving heating ice from -10°C to -1°C and then converting it to water. The specific heat of ice is established as half of that of water, which is 4.2 kJ/(kg·°C), leading to a value of 2.1 kJ/(kg·°C) for ice. The correct approach to the problem involves using the equations Q = mcT and Q = mL, with the realization that the heat of fusion must be included in the energy calculations when transitioning from ice to water.

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Homework Statement



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.


Homework Equations


Q=mct
Q=mL


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 )
 
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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
 
Thanks a lot Tusike .
 

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