Please help on chemistry energy problem please?

Question

Could you please help me on my chemistry energy problem
it's so hard
i have answer from the teacher
but i don't get how to work this problem out
could you please show work
thanks ^^

2. A 752 cubic centimeter sample of water was placed in a 1.00 kg aluminum pan. The initial temperature of the pan was 26 Degrees Celcius and the final temperature of the system was 39 Degrees Celcius. What was the initial temperature of the water?

answer = 43 Degrees Celcius

Answer 1

E - Energy.
C - Specific heat capacity.
Mass.
temp. ▲ - Temperature change.
Specific heat capacity of Water : 4180 J kg^-1 k^-1
Specific heat capacity of Aluminum : 9∙1 x 10² J kg^-1 k^-1

Assuming all of the heat energy came from the water and none was lost.

Heat energy received from the water to the Aluminum:
E = C x Mass x temp. ▲
E = (9∙1 x 10²) x (1∙0) x (39°- 26°)
E = 11830 Joules.

So 11830 Joules was taken from the water to heat the Aluminum pan. How many degrees will the water raise by, if this energy was applied to it?

E = C x Mass x temp. ▲
temp. ▲ = E / [C x Mass]
temp. ▲ = 11830 / [4180 x 0∙752]
temp. ▲ = 11830 / 3143.36
temp. ▲ = 3∙763 488 751...º
So the temperature of the water dropped by 3∙763 488 751...º when it heated the Aluminum pan.

Initial temperature of water:
Initial temp. = Present temp. + Drop in temp.
Initial temp. = 39 + 3∙763 488 751...º
Initial temp. = 42∙763 488 751...º
Initial temp. ≈ 43º

Answer 2

First of all we assume that during the pouring of water no heat will be lost (which is not true but impossible to estimate without knowing more) Then we find the heat capacities Cp of water and aluminium.
Cp water =4,18kJ/kg*deg.C
Cp aluminium =0,9kJ/kg*deg.C
To heat up aluminium from 26->39 deg.C: 0.9 * 13 deg.C * 1kg =11,7kJ heat is needed. This energy comes from the water that is cooling down from the unknown
temperature X to 39 deg.C.
X-39 = 11.7 kJ / (4.18 * 0.752) = 3,73 deg.C
From this it would follow that the water = 39 + 3,73 deg.C = 42,73 deg.C
The volumetric coefficient of expansion of water is 210E-06 /deg.C. This means that at 39 deg.c the density of water is 0,991 kg/m3. At 43 deg.C 752 cm3 water will actually weigh 0,745 kg
X is therefore 39+11.7 / (4,18 * ,745) = 42,76 deg.C or if ronded off 43 deg.C