General Chemistry Homework Question

Question

Suppose that 4535.0 J of heat energy is added to 9 g of ice initially at -8 C. (This could be done, for example, by electrical heating of ice in an insulated container.) What is the final temperature of the water?



Properties of Water
CP(solid)	1.96 J/g/K
ΔHfus	6.02 kJ/mol
CP(liquid)	4.184 J/g/K

Answer 1

Initially the solid is heated from - 8 C to zero C where the solid phase go through a phase transition to the liquid phase. Then this liquid is heated with the remaining energy to the final temperature.

The change in temperature of the solid phase is 8 C with heat input of 1.96 J /(g-K) x 9 grams x 8 K

Ball park this is about 144 Joules of heat a bit less so now we have about 4391 Joules for making the phase change.....4400 Joules

Water has a molecular weight 18 grams per mole so 9 grams is 1/2 a mole.

the energy required to melt the 9 grams of ice is a little over 3,000 Joules for 1/2 a mole.

So this leaves about 1,400 Joules for heating the liquid.

Then we take 4.184 J/(g-K) x 9 grams and divide this into 1, 400 J to obtain the ball park temperature of the water. 1,400 J/ 40 J/K is about 35 K or 35 C to one sig figure 4 x 10 C

Now lets look at exact numbers:

Energy input to raise temp of ice is 141.12 J

Energy input to change phase is 3.01 KJ

Remaining energy for liquid temperature change is now 1,383.88 Joules

divide by 4.184 x 9 = 37.657 J/K

Yields 36.74961 K

Mass is given to 1 sig fig so we should expect a final temperature of 40 C or really 4 x 10¹ C