- #1
thunder
- 22
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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!
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!