Calorimeter Constant problem help

[AwesomOne]

I am stuck on how to do this problem:

       A calorimeter is to be calibrated 51.203 g of water at 55.2 degrees C is added to a calorimeter containing 49.783g of water at 23.5 degress C. After stirring and waiting for the system to equilibrate, the final temperature reached is 37.6 degrees C. Calculate the calorimeter constant.

I think that you have to 55.2- 37.6 and 37.6- 23.5  and I also know that you need the 4.184 constant for specific heat of water... I am stuck what to do next.
Thanks         

[kingchemist]

I assume you will have to find the temperature of mixing the two sample of water, assuming no heat is absorbed by the calorimeter. I expect the temperature will be >37.6 degrees C. The difference in temp will be the result of heat being absorbed by the calorimeter.
Let the temperature of water on mixing be w degrees

As c is constant, then m1[delta]T = m2[delta]T
                              51.203 x (55.2-w) = 49.783 x (23.5 + w)

Solve for w.

Find q at this temperature using g = cm[delta]T where m=the combined mass of water and c is the specific heat of water in the right units.
Find q at 37.6 degrees C and the difference in the two values is the amount of heat absorbed by the calorimeter.

I expect the constant will be this difference in the amount of heat/the difference in temperature and so will have units of J (or kJ) per degree C

[valdorod]

q_hot = -(q_cold+q_cal)

q_hot = m_hot*s_water*ΔT_hot = 51.206*4.184*(-17.6)

q_cold = m_cold*s_water*ΔT_cold = 49.783*4.184*14.1

q_cal) = C*ΔT_cold = C*14.1

plug in and solve for C