[burtsbee]

Hi,

I can't find any worked examples of this type of problem on the internet that include the specific heat capacity of the calorimeter:

If 60g of Al metal at 100ºC were added to 60g water at 18ºC in a calorimeter of heat capacity (C) 30J/K. what would the final common temperature be, assuming no heat losses? (C(Al) = 0.897 J/g.K and C(H2O) = 4.184J/g.K)

So my working out so far is:

heat lost Al = heat gained H2O + heat gained calorimeter

Q=mc x change in temp

60 x 0.0897 x (100 - Tf) = 60 x 4.184 x (Tf - 18) + 30 x (Tf - 18)

- cross out both the 60g.

89.7 - 0.897Tf = 4.18Tf + 30Tf - 540
705.012 = 35.059Tf
Tf = 20.1 = 20ºC

But I seriously doubt the final temperature is 20ºC.I am also confused over if I have to convert ºC into K for this, because none of the other worked problems have converted.

Any help is much appreciated.

[Yggdrasil]

You can't cancel out the 60g from both sides because the 60g does not occur in the expression for the heat gained by the calorimeter.

[burtsbee]

Even if I leave the 60g on both sides it only changes my answer to 19.5degrees.

[Borek]

Check your math, that's not what I got.