What is the Total Heat Capacity of the Calorimeter?

[gogeta2006]

Homework Statement

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.

Homework Equations

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

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.
Please someone please show me how to correct this.

Thank you.

 

[Astronuc]

Heat absorbed by calorimeter = 250 + 840 = 1090 cal

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)

 

[gogeta2006]

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)

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

 

[Borek]

Show your detailed work then.