1. Whatever heat is supplied to the system is utilised in 4 processes, namely
To increase temperature of ice
To increase temperature of water.
To increase temperature of cup
To change the state of ice to water
So total heat given = q(cup) + q(water) + q(ice) + heat of fusion
2. q(water) is the amoutaof heat the water molecules in liquid phase contains.
q(ice) is the heat content of water molecules in ice
q(cup) is the heat content of molecules of cup
Heat of fusion of ice is the amount of heat that is required to change the state of water from solid to liquid at a constant temperature.
3. Specific heat is the heat required to raise the temperature of one gram of a system by one degree.
So, considering the temperature to be changed is 20degrees . q(water) = specific heat * change in temperature.
= 4.18 * 20 = 83.6 joules for one gram of water.
4. In the same way as in part 3. Only by using the specific heat of ice.
5. While measuring the q(cup) we should know the temperature change of the cup. And in the same way as above we can find it using specific heat of cup molecules.
6. If we know the amount of heat we have supplied to the system and we can measure the temperature change of specifically water, ice and cup each, we can find the unknown quantity,that is heat of fusion of ice.
7. This is because as the amount of ice is increased the amount of heat needed is also increased. So that would not give us a standard value of heat of fusion . So in order to avoid such error we divide the value by no.of moles. In this way we get a standard value which will be true in all cases irrespective of the amount of ice taken.
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