Need help with simple calorimetry problem

[kishin7]

So here's the problem

Conside the dissolution of CaCl2:

CaCl2(s) ---> Ca^(2+) (aq) + 2Cl^(-1) (aq) deltaH = -81.5kJ

An 11.0g sample of CACl2 id dissolved in 125g of water, with both substances at 25.0°C. Calculate the final temperature of the solution assuming no heat lost to the surroundings and assuming the solution has a specific heat capacity of 4.18 J/(°C*g).


So far I think it's just q(rxn) = q(water) + q(CaCl2)

(136g)(4.184J/°Cg)(Tfinal - 25.0°C) = (125g)(4.184)(Tfinal - 25.0°c) + (11g)c(Tfinal - 25.0°C)

But it seems that I am missing C for the calcium chloride and I don't know what to do about it? is there another way to calculating it without C of CaCl2?

 

[so-crates]

The heat capacity(Cp) of the solution will be very close to that of pure water, 4.18 J/(g C). The problem states this.

 

[kishin7]

Are you suggesting that I use 4.184 for the heat capacity of every one? If I do then it wouldn't make sense since that's just making the left side equal to the right side and getting at final temperature of 25.0°C.

By the way, is my formula correct though?

heat of solution = heat of water + heat of calcium chloride

 

[GCT]

So far I think it's just q(rxn) = q(water) + q(CaCl2)

(136g)(4.184J/°Cg)(Tfinal - 25.0°C) = (125g)(4.184)(Tfinal - 25.0°c) + (11g)c(Tfinal - 25.0°C)

You won't need the left side of this equation. Note that the enthalpy is in kJ/mole of reaction. How would you convert kJ/mole of reaction to kJ?

This energy is equivalent to q...it is used to raise the temperature of ___?

"(136g)(4.184J/°Cg)(Tfinal - 25.0°C)",

this is incorrect, 4.184J/Cg applies specifically to water.