# Calculate Total Heat Capacity

1. Mar 11, 2009

### gogeta2006

1. The problem statement, all variables and given/known data

50ml of water at 49.6 C were mixed with 50ml of water at 25.1 C in a calorimeter also at 25.1 C. The final temperature was 30.1 C Assuming that neither the density of water nor its specific heat capacity change with temperature, calculate the total heat capacity of the calorimeter.

2. Relevant equations

Density of water = 1.00 g/mL
Specific heat capacity = 4.18 J / g * K

3. The attempt at a solution

q (heat given up by water) = 50ml * (49.6-30.1)
= 840 cal

q (heat absorbed by cold water) = 50ml (30.1-25.1)
= 250 cal

Heat absorbed by calorimeter = 250 + 840 = 1090 cal

Ccal = qcal / delta T
= 590 / (30.1-25.1)
= 118 K

The answer is supposed to be 493.24 J/K .... but i am not getting that.

Thank you.

2. Mar 11, 2009

### Astronuc

Staff Emeritus
Woah!

Heat is lost from the hot water to the calorimeter AND the cold water. One cannot at the heat of the hot and cold water.

Try ΔQ(hot) = ΔQ(calorimeter) + ΔQ(cold)

3. Mar 11, 2009

### gogeta2006

I tried doing that and the answer is 118 (which is still INCORRECT)...i wrote plus there where it should be minus.

Last edited: Mar 11, 2009
4. Mar 12, 2009