Specific Heat of Metals/Homework Help?

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The equilibrium temperature of a mixture of 50 cm³ of water at 85 degrees Celsius and 100 cm³ of water at 10 degrees Celsius is calculated using the formula for heat transfer. The specific heat of water is 4184 J/kg°C. The correct setup leads to an equilibrium temperature of 35 degrees Celsius. The discussion clarifies that temperature can be expressed in Celsius for this type of problem, negating the need for absolute temperature conversions.

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thunder
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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!
 
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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.)
 

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