How Do You Calculate the Specific Heat Capacity of an Unknown Substance?

[physicsnerd7]

Homework Statement

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.

Homework Equations

Q_released + Q_absorbed = 0
Q = mcΔT

The Attempt at a Solution

So i did:
-Q_{unknown substance (us)} = Q_water + Q_cup

- m_usc_us (T_{f us} - T_{i us}) = mw_w c_w (T_fw- T_iw )+ m_c c_c (T_fc-T_ic)

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

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

c_s= 11345.5 / 5.51

c_s = 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 Q_absorbed so i got this :

- m_usc_us (T_{f us} - T_{i us}) = mw_w c_w (T_fw- T_iw )

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

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

c_s= 1715.06J/kgC

I believe this answer is wrong! but i don't know!

 

[CWatters]

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