# Calorimeter Cup Problem!

1. Dec 4, 2013

### physicsnerd7

1. The problem statement, all variables and given/known data
A copper calorimeter cup with a mass of 0.27 kg contains 0.125 kg of water. At 17°C. If 95g of an unknown substance at 93°C is placed into the cup, the temperature increases to 35°C. Find the specific heat capacity of the unknown substance.

2. Relevant equations
Qreleased + Qabsorbed = 0
Q = mcΔT

3. The attempt at a solution
So i did:
-Qunknown substance (us) = Qwater + Qcup

- muscus (Tf us - Ti us) = mww cw (Tfw- Tiw )+ mc cc (Tfc-Tic)

-(0.095kg) cus (35 C - 93C) = (0.125 kg) (4.2E3 J/kgC) (35 C -17C) + (0.27 kg)(3.9E2 J/kgC) (35C-17C)

cs = ((0.125 kg) (4.2E3 J/kgC) (35 C -17C) + (0.27 kg)(3.9E2 J/kgC) (35C-17C)) / -(0.095kg)(35 C - 93C)

cs= 11345.5 / 5.51

cs = 2059 J/kgC

Okay so that is my answer (2059 J/kgC)and I would just like clarification if my answer is right because before i did it a different way and didnt inlcude the copper cup in the Qabsorbed so i got this :

- muscus (Tf us - Ti us) = mww cw (Tfw- Tiw )

-(0.095kg) cus (35 C - 93C) = (0.125 kg) (4.2E3 J/kgC) (35 C -17C)

cs= ((0.125 kg) (4.2E3 J/kgC) (35 C -17C)) / -(0.095kg)(35 C - 93C)

cs= 1715.06J/kgC

I believe this answer is wrong!! but i dont know!

2. Dec 5, 2013

### CWatters

I haven't checked your answer but it's reasonable to include the effect of the copper cup.