Specific Heat of Metals/Homework Help?

[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!

 

[berkeman]

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.)

 

[kyle8921]

I can confirm that your approach and calculations for this problem are correct. The formula you used is the specific heat equation, which is commonly used to calculate the equilibrium temperature when two substances at different temperatures are mixed together. The specific heat of water (c) is indeed 4184 J, and using the given values for mass and temperature, your final answer of 35 degrees C is correct. However, it is always a good practice to double check your calculations and make sure all units are consistent throughout the calculation. Keep up the good work!