Calorimeter Constant Homework - Calculate 330 J/C

[Grunstadt]

Homework Statement

Calculate the calorimeter constant
50 ml H2O = 50g
Starting Temp for hot water = 52.3°C
Starting Temp for cold water = 21.8°C
Final Temp = 41.3°C
Change in Temp = Hot = -11°C / Cold = 19.5°C

Homework Equations

Q = mcΔT

The Attempt at a Solution

Q = (50g)(4.184)(11) = 2301 J
Q = (50g)(4.184(-19.5) = -4079 J

2301 - (-4079) = 6380 J

6380J / 19.5 = 330 J/C

Now my main questions is since the cold water is gaining heat would the joules be negative or positive.

 

[Lamebert]

The transfer of heat was from the warm water to the cold water, so the heat transfer of energy of the cold water is positive.