Problem — Water and oil at different temperatures

Problem

A container holds $2\;\mathrm{kg}$ of water at $T_1 = 293\;\mathrm{K}$ ($c_{s,1} = 4180\;\mathrm{J/(kg\,K)}$). A second container holds $1\;\mathrm{kg}$ of oil at $T_2 = 423\;\mathrm{K}$ ($c_{s,2} = 2090\;\mathrm{J/(kg\,K)}$). Bringing the two containers into thermal contact (isolated system), find the final equilibrium temperature.

Solution

Set-up. The system is isolated: the thermal energy given up by the oil (hotter) is absorbed by the water (colder). At equilibrium both reach the same temperature $T_f$. The calorimetric balance requires the sum of the thermal energy changes to be zero:

$m_1 c_{s,1}\,(T_f - T_1) + m_2 c_{s,2}\,(T_f - T_2) = 0$

Symbolic solution for $T_f$:

$T_f = \frac{m_1 c_{s,1} T_1 + m_2 c_{s,2} T_2}{m_1 c_{s,1} + m_2 c_{s,2}}$

Heat capacities of the two bodies:

$m_1 c_{s,1} = 2\cdot 4180 = 8360\;\mathrm{J/K}, \qquad m_2 c_{s,2} = 1\cdot 2090 = 2090\;\mathrm{J/K}$

Numerical substitution:

$T_f = \frac{8360\cdot 293 + 2090\cdot 423}{8360 + 2090} = \frac{2\,449\,480 + 884\,070}{10\,450} = \frac{3\,333\,550}{10\,450}$

$\ev{T_f \approx 319\;\mathrm{K} \approx 46\;^\circ\mathrm{C}}$

The equilibrium temperature is much closer to that of the water than that of the oil: the water has four times the heat capacity (double the mass and double the specific heat), so it “pulls” the equilibrium towards itself.

Collegamenti

Argomenti: Work and energy Concetti: Specific heat and heat capacity · Thermal energy from friction · Conservation of mechanical energy Competenze: Calorimetric equation · Symbolic set-up