# Temperature Changes

## Homework Statement

If 25g of water at a temperature of 10 degrees Celsius is mixed with 40 grams of water at 80 degrees Celsius what is the final temperature of the mixture?

## Homework Equations

Qlost + Qgained = 0
Q=mcΔt

## The Attempt at a Solution

Attempt 2:
0.025(4186)(x-10) + (0.04)(4186)(x-40) = 0
x = 28.46

I don't know if I'm doing this right...

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I don't know if I'm doing this right...
You have the right idea, just remember that delta T = Tf - Ti

So the water at 10 degrees is going to heat to some final temperature T, and the water at 80 degrees is going to cool to the same temperature T

we don't need the specific heat of water since it's common in both terms

(0.025)(10 - x) = (0.04)(x-80)

x = 53 degrees

Also note in your equation - your heat is 40 degrees rather than what the question states - 80 degrees. Perhaps you were thinking of 40g

hope this helps,

cheers

Thank you so much!