Specific heat capacity of calorimeter

[Pouyan]

I've got a problem:
A piece of copper with mass m1 = 800 g and temperature t1 = 80 ° C is placed in a container with good thermal insulation. The vessel initially contains water with mass m2 = 500 g temperature t2 = 20 C. What is the calorimeter (including thermo meter) heat capacity if the end temperature is tf = 26 C?!

The solution is:

the specific heat capasity for copper: 0:39 kJ / kg.C
and for water: 18.4 kJ / kg.C
dQ copper = m1 * CCU * (t1-t0) = 16.85 kJ

dQ water = m2* CH20 * (t0-t1) = 12:54 kJ

According to the first law of thermodynamics, the amount of heat transferred to the calorimeter and thermometer:

dQ = dQ copper - dQ water = 4.31 kJ

But how can I find the heat capacity of the calorimeter?
I see a solution that I should do this: 4.31 kJ / (26-20) C

but why water in this case?! should I always find the specific heat capacity of the calorimeter or the thermometer with respect to the minimum tempraturen?!

 

[Korak Biswas]

Heat capacity(not specific heat) is nothing but mass times specific heat. If h be the heat capacity of the calorimeter including thermometer and your calculation is correct, then
$h(26-20) = 4.31$

 

[Pouyan]

Heat capacity(not specific heat) is nothing but mass times specific heat. If h be the heat capacity of the calorimeter including thermometer and your calculation is correct, then
$h(26-20) = 4.31$

Should I think:

Q los = Q gain

and Q gain = Q water + Q calorimeter ?!

 

[Korak Biswas]

Should I think:

Q los = Q gain

and Q gain = Q water + Q calorimeter ?!

Heat emitted by piece of copper = heat absorbed by water + heat absorbed by calorimeter including thermometer.
It follows from the conservation of energy.

 

[Pouyan]

Heat emitted by piece of copper = heat absorbed by water + heat absorbed by calorimeter including thermometer.
It follows from the conservation of energy.

So we reduce 26 with the temprature of water for calorimeter ...is this always like this? If water is in the system should we find always the heat capacity of calorimeter with respect to water temprature ? What if water is not involved ?!I ask for an apology that I asked like this but I am a beginner...
thanks !

 

[TSny]

In a problem like this, it is assumed that the water and the calorimeter are initially in thermal equilibrium with each other before the copper is added. So, you can assume the calorimeter and the water have the same initial temperature (20 ^oC in this case.)

If there is no water so that the copper is placed in an empty calorimeter, then you would need to be given the initial temperature of the calorimeter.

 

[Korak Biswas]

So we reduce 26 with the temprature of water for calorimeter ...is this always like this? If water is in the system should we find always the heat capacity of calorimeter with respect to water temprature ? What if water is not involved ?!I ask for an apology that I asked like this but I am a beginner...
thanks !

I think now you will have a clear idea after going through TSny's reply.