Worked exercise — Calorimeter of mixtures

Problem

In an ideal calorimeter (no losses) we pour $300$ g of water at $80\ ^{\circ}$C and $200$ g of water at $20\ ^{\circ}$C. What is the equilibrium temperature?

Solution

Setup. Isolated system: the heat released by the hot water equals the heat absorbed by the cold water. With the same specific heat $c$: $m_1 c (T_e - 80) + m_2 c (T_e - 20) = 0$ Substitution ($c$ cancels out, masses in grams): $300\,(T_e - 80) + 200\,(T_e - 20) = 0$ $500\,T_e = 28000$ $\ev{T_e = 56\ ^{\circ}\mathrm{C}}$ It is the mass-weighted average: $0{,}6\cdot 80 + 0{,}4\cdot 20 = 56$.

Collegamenti

Argomenti: Termologia Competenze: Equazione calorimetrica Metodi: Equazione calorimetrica delle miscele Oggetti: Calorimetro