# Specific Heat of Metals/Homework Help?

1. Apr 20, 2006

### thunder

Here is a homework problem I am working on:

1. Suppose 50 cm^3 of water at 85 degrees C is addedto 100 cm^3 of water at 10 degrees C. What is the equilibrium temperature of the mixture, assuming there is no heat lost to the environment?

* = multiplied by
T = Temperature of water
m = mass of water
c = specific heat of water = 4184 J

T= {(m * c * T) + (m * c * T)} all that divided by {(m * c + m * c)}

which gives me the following:

{(50 * 4184 * 85) + (100 * 4184 * 10)}
------------divided by---------------
{(50 * 4184) + (100 * 4184)}

= 35 degrees C for equilibrium temperature

Is this correct? Did I set it up and do it the right way??

Appreciate your help on this :) THANKS!!

2. Apr 20, 2006

### Staff: Mentor

Is the temperature supposed to be in C or K? (I honestly don't know, since I don't work with this type of problem, but it seems like absolute temperature might be needed for the calculation of the combined average, and then convert back to C.)