(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant 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

3. 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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# Homework Help: Specific Heat Capacity question

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