Mixing two batches of water

Problem

$200\;\text{g}$ of water at $80\;^\circ\text{C}$ are mixed with $300\;\text{g}$ of water at $10\;^\circ\text{C}$. Neglecting losses to the surroundings, what is the final temperature $T_f$?

Solution

Setup. At thermal equilibrium the heat given up by the hot water equals the heat absorbed by the cold water: the algebraic sum of the heat exchanged is zero. $m_1 c\,(T_f - 80) + m_2 c\,(T_f - 10) = 0$ Since both are water, the specific heat $c$ cancels: the result is the weighted average of the temperatures with the masses.

Numerical calculation. $T_f = \frac{m_1 \cdot 80 + m_2 \cdot 10}{m_1 + m_2} = \frac{200\cdot 80 + 300\cdot 10}{200 + 300} = \frac{19000}{500} = 38\;^\circ\text{C}$

$\ev{T_f \approx 38\;^\circ\text{C}}$

The value falls between $10$ and $80\;^\circ\text{C}$ and is closer to $10$ because the cold mass is larger: consistent.

Links

Topics: Thermodynamics Concepts: Specific heat and heat capacity Skills: Calorimetric equation Methods: Calorimetric equation for mixtures Objects: Calorimeter