# Specific Heat Capacity question

## Homework Statement

If 100 mL test of water at 60°C is added to 100 mL of water at 20°C, then the final temperature of the water will be?

## Homework Equations

c=Specific Heat Capacity

Specific Heat Capacity of water = 4186

Quantity of Heat Energy=mass * Specific Heat Capacity of substance * temperature difference
-or-
Q-=m*c*temp diff

100mL=0.1 kg

## The Attempt at a Solution

Water quantity at 60 degrees + Water quantity at 20 degrees = sum water quantity at new temperature

0.1 kg * 4186 * (60 - final temp) + 0.1 kg * 4186 * (final temp - 20) = 0.2 kg * 4186 * final temp

418.6 (60-final temp) + 418.6 (final temp-20) = 837.2 * final temp
25116 - 418.6final temp + 418.6final temp - 8372 = 837.2final temp
16744=837.2final temp
final temp= 20 degrees celsius

So, the final temp is 20°C ??? that doesnt make sense?

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Oh hold on now. I think I see my mistake. I didn't account for the Law of Conservation of Energy.

Final Heat Change of 60°C Water + Final Heat Change of 20°C Water = 0

0.1 kg * 4186 * (final temp-60) + 0.1 kg * 4186 * (final temp - 20)=0

418.6 (final temp) - 25116 + 418.6 (final temp) - 8372 =0

837.2 (final temp) = 33488

final temp = 40

Therefore, the final temperature of the water is 40 °C

Ygggdrasil