Physics help? (heat capacity and calorimeter questions)

Here are the two questions (this was actually from a test, and my teacher marked them wrong but i thought i got them right.)

1. 950 g of a liquid requires 3800 J of energy to raise the temperature by 3 degrees celsius. What is the heat capacity of the sample?
(they ask for specific heat capacity in the next question)

My attempt:
because its only heat capacity, not specific heat capacity, the formula is:

Heat capacity=Energy/change in temperature
=3800/3=1267 J/K. My teacher marked this wrong. (however, the specific heat capacity is 1333, which he marked right).

2. A 21.0 g sample of water at 3 Celsius is cooled down to transform it into ice at -3 Celsius. The amount of heat withdrawn from the sample is: a) 7195, b) 6930, c) 7327, d) 264 (all in Joules)

My attempt:
There are three heats: the heat withdrawn from the water while cooling from 3 celsius to 0 celsius,
the latent heat of fusion as the water transforms to ice,
the heat withdrawn from the ice while cooling from 0 to -3 Celsius.

Therefore, the total heat (E) should equal:

E = m(water) * spec. heat capacity (water) * ΔT (water) + m(water) * lat. heat of fusion (water) + m (ice) * spec. heat capacity (ice) * ΔT (ice)

or: MC(water)ΔT + ML + MC((ice)ΔT


= 21.0*4.18*(0-3) + 21*334 + 21*2.2* (-3-0)
=6612 J
The reason i put (0-3) and (-3-0) is because ΔT is Tfinal-Tinitial, and those are the respective final and initial temperatures.

However, 6612 J isn't really close enough to any of the multiple choice answers. Can anybody help me?

Thanks a lot!

Q1. Your are correct for both parts of Q1. Your teacher got it wrong.

Q2 You have mis-used the 'signs'. When you say 'The reason i put (0-3) and (-3-0) is because ΔT is Tfinal-Tinitial, and those are the respective final and initial temperatures.' this leads to a problem. Your approach gives negative values for the 3degrees drops; this is OK if you remember to make the latent heat ALSO negative (use -21.0x334 not +21.0x334). So then all of your values will be negative, corresponding to the removal of heat.