Heat/First Law of Thermodynamics/Calorimetry - Have work done

[maniacp08]

Heat/First Law of Thermodynamics/Calorimetry -- Have work done

A calorimeter of negligible mass contains 920 g of water at 303 K and 48 g of ice at 273 K. Find the final temperature T.
°C
Solve the same problem if the mass of ice is 460 g.
°C

Formula I've used
Mice * Lf + Mice * Cwater * change in T icewater = Mwater * Cwater * change in T water

(.048kg)(79.7kcal/kg)+(.048kg)(1kcal/kgxK)(Tf - 273K) =
(.92kg)(1kcal/kgxK)(303K-Tf)

Ive gotten Tf to be 297.56K which is 24.41 C

and for part B I've gotten the answer to be 266.433K which is -6.7167 C

Is this correct?

 

[mburt3]

I would start with the idea that the -q(water)= q(ice)
This is because the water is warmer and it is losing heat to the ice. It will continue transferring heat until both are at the same temperature. Since the calorimeter has negligible heat capacity, it does not absorb any of the heat that the water gives off.
From here you can say that the -ms(t(fin)-t(int)) for water is equal to ms(t(fin)-t(int)) for the ice.
Hope that helps.

 

[thomate1]

I cannot confirm if your calculations are correct without seeing your work and the specific values for the specific heat and latent heat of fusion used. However, your approach and use of the first law of thermodynamics and calorimetry principles seem correct. It is always important to double-check your calculations and make sure the units are consistent. Additionally, if possible, it is always beneficial to compare your results with known values or to repeat the experiment to confirm your findings.