Calorimetry

Question

A 50 g sample of water at 20°C is mixed with 50 g of water at 80°C. The final temperature of the mixture is:

RankGuru · Accepted Answer

Assume there is no heat loss to the surroundings. 
$Heat \, Lost \, by \, Hot \, Water = Heat \, Gained \, by \, Cold \, Water$
$m_1 c_p (T_{hot} - T_f) = m_2 c_p (T_f - T_{cold})$
- $m_1 = m_2 = 50 	ext{ g}$
- $c_p$ cancels out.
$80 - T_f = T_f - 20$
$2T_f = 100 \implies T_f = 50^\circ C$
- Since masses and substances are identical, the final temperature is the simple arithmetic mean.

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