(adsbygoogle = window.adsbygoogle || []).push({}); 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.

Please someone please show me how to correct this.

Thank you.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Calculate Total Heat Capacity

**Physics Forums | Science Articles, Homework Help, Discussion**