Worked exercise — Heat exchange between three fluids

Problem

In a perfectly insulated calorimeter, three substances are mixed: $m_1 = 100 \ \text{g}$ of water at $80 \ ^{\circ}\text{C}$, $m_2 = 200 \ \text{g}$ of oil ($c_2 = 2000 \ \text{J/(kg·K)}$) at $20 \ ^{\circ}\text{C}$ and $m_3 = 50 \ \text{g}$ of copper ($c_3 = 385 \ \text{J/(kg·K)}$) at $5 \ ^{\circ}\text{C}$. Determine the equilibrium temperature, assuming no substance reaches boiling point. (Assume $c_{\text{acqua}} = 4186 \ \text{J/(kg·K)}$.)

Solution

In an isolated system the sum of the heats exchanged is zero, $\sum_i m_i c_i (T_e - T_i) = 0$, where the hotter substances release heat (negative contribution) and the colder ones absorb it. Let us write the calorimetric equation with masses in kg: $0{,}1 \cdot 4186 \cdot (T_e - 80) + 0{,}2 \cdot 2000 \cdot (T_e - 20) + 0{,}05 \cdot 385 \cdot (T_e - 5) = 0$ Expanding the products: $418{,}6 \, T_e - 33488 + 400 \, T_e - 8000 + 19{,}25 \, T_e - 96{,}25 = 0$ Collecting the coefficients of $T_e$ and the known terms: $837{,}85 \, T_e = 41584{,}25 \implies T_e = \frac{41584{,}25}{837{,}85} \approx 49{,}6 \ ^{\circ}\text{C}$ Water, despite having a smaller mass than the oil, dominates the result thanks to its more-than-double specific heat. $\ev{T_e \approx 49{,}6 \ ^{\circ}\text{C}}$

Collegamenti

Argomenti: Termologia Concetti: Calore specifico e capacità termica Competenze: Equazione calorimetrica Metodi: Equazione calorimetrica delle miscele Oggetti: Calorimetro