[Cyberconvict]

I'm a bit unsure about this process and just want to know if I'm doing the right thing. Here's the problem:

A coffee-cup calorimeter contains 150.0g water at 25.1 C. A 121.0 g block of copper is heated to 100.4C. The specific heat of Cu(s) is 0.385 J/g-K. The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant temperature of 30.1C.

(A) Determine the amount of heat, in J, lost by the copper block.
q = -(121.1g)(0.385J/g-K)(100.4-30.1C) = -3274.9 J

(B) Determine the amount of heat gained by the water.
q = -(150g)(4.18J/g-K)(25.1-30.1C) = 3135 J

(C) The difference between answers (a) and (b) is due to heat loss through the Styrofoam cups and the heat necessary to raise the temperature of the inner wall of the apparatus.  Calculate the heat capacity of the calorimeter in J/K.

3274.9 J - 3135 J = 139.9 J
139.9 J/5K = 27.9 J/K

(D) Doesn't matter just asked for a verbal explanation.

(E) What would be the final temperature of the system if all the heat lost by the copper block were absorbed by the water in the calorimeter?

q Cu = q H2O I'm assuming is correct.

so,

(121g)(0.385 J/g-K)(100.4C-x) = (150g)(4.18 J/g-K)(x-25.1C)
x= 30.3C

Thank you in advance. If it's all correct just tell me and don't worry about explaining in any more depth.

[xiankai]

121.1g

typo?

trivial thing but just so u note: there's no need for the negative sign as they already mentioned the word "loss"  

150g

try to write it as 150.0; i dont know if your teachers are strict but they insist on having the correct number of sig. fig. / d.p. and will penalise you for incorrect usage.

139.9 J/5K = 27.9 J/K

since u rounded to 3 sig fig. it should be 28.0 J/K rather.

OK, all your answers seem right o me, good work