Question

Determination Of The Molar Heat of Fusion Of Ice Purpose The purpose of this result with its reported value that is s 6.01 kJ/mol. The molar heat of fusion (AH) of ice is the amount of heat required to melt 1 mole of ice experiment is to determine the molar heat of fusion of ice and compare the experimental Definition Heat of fusion of ice is the amount of heat required to melt ice. The more the ice to melt, the more the heat is obviously required. This means that we need to normalize (or standardize) the heat of fusion of ice, that you will determine in this experiment in the unit of J or kJ, by dividing it by the number of moles of the ice you ill use. This normalization finally gives you the molar heat of fusion of ice in the unit J/imol or kJ/imol Why Does Heat of Fusion of Ice (or of any solid) Exist? How much heat needs to be supplied to change the temperature of 100 gram ice at Using the specific 0 °C to 20 °C water? heat of water 4.18 J/gC, you may calculate the amount of heat required as follows: The amount of heat needed = 4.18 J/g°C x 100 g x (20-0)"C-836 J This result, however, is an underestimate. In other words, if you supply the heat of 836 J, the temperature will not reach 20 °C but fall far short from it. This is because most of the heat supplied is used to reorient (or push apart) water molecules in ice rather than to increase the temperature. Since temperature is a measure f molecular motion, water molecules that are stuck in ice must get loosened before they start moving around, which will eventually result in temperature increase as freed water molecules move faster and faster as they receive more and more heat. This portion of heat that is spent to restructuring the ice is the heat of fusion of ice. There is no temperature increase over the period over which the heat of fusion is used. Temperature increases only after the heat of fusion is all consumed. (Refer to Figure 11.38 in the textbook) By incorporating the heat of fusion of ice, you can make the correct calculation to this question as follows: The amount ofheat needed = 6.01 kJ/mol x (20 g / l 8 g/mol) + 4.18 J/g°C x 100 g x (20-0)"C 7513.78 J Thus, this correct amount of heat needed, 7513.78 J, is much greater than the incorrect amount of 836 J Experimental Setting The following analogy should help setting up an experi ment for you to measure the heat of fusion of ice PRE-LABORATORY QUESTIONS 1. Set-up the mathematical relation that should goven the heat transfer processes in this expeniment o el Name: using such terms as qwater, Qice, qcup, and the heat of fusion of ice. 2. Classify the four terms (Qwater, qies, Qcup, and the heat of fusion of ice) into two groups: the heat gaining group and the heat losing group. 3. Describe how you measure qwater, the amount of heat the water (loses? Or gains?). You are to use the specific heat of water, 4.18 J/g C 4. Describe how you measure qice, the amount of heat the ice (loses? Or gains?). You are to use the specific heat of ice, 2.06 J/g°C. 5. Describe how you measure qcup, the amount of heat the styrofoam cup (loses? Or gains?). You are to use the heat capacity of the styrofoam cup, 10 J/C. Describe how you finally determine the heat of fusion of ice. on of ice you determine in the previous question is either J (joules) or kJ (kilo pules). Explain why we need to divide this heat of fusion of ice by the number of moles of the ice you ill use so that the final unit should be J/mol or kJ/mol the unit for the molar heat of fusion of ice.

Answer 1

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.