Problem — Volcanic rock in Fanta

Problem

A volcanic rock ($4\;\mathrm{kg}$, $c_s = 300\;\mathrm{J/(kg\cdot K)}$) at $400\;^\circ\mathrm{C}$ is dropped into $8$ litres of Fanta ($c_s = 30\;\mathrm{J/(kg\cdot K)}$, density $1{,}5\;\mathrm{kg/l}$) at $-50\;^\circ\mathrm{C}$. Find the equilibrium temperature. (Careful: work in kelvin.)

Solution

Conversion to kelvin (essential for the thermal energy $E = m\,c_s\,T_K$):

$T_{1} = 400 + 273 = 673\;\mathrm{K} \quad(\text{rock}), \qquad T_{2} = -50 + 273 = 223\;\mathrm{K} \quad(\text{Fanta})$

Mass of the Fanta from volume and density:

$m_2 = 8\;\mathrm{l} \cdot 1{,}5\;\mathrm{kg/l} = 12\;\mathrm{kg}$

Calorimetric balance (isolated system, thermal energy conserved):

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

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

Heat capacities:

$m_1 c_{s,1} = 4\cdot 300 = 1200\;\mathrm{J/K}, \qquad m_2 c_{s,2} = 12\cdot 30 = 360\;\mathrm{J/K}$

Numerical substitution:

$T_f = \frac{1200\cdot 673 + 360\cdot 223}{1200 + 360} = \frac{807\,600 + 80\,280}{1560} = \frac{887\,880}{1560}$

$\ev{T_f \approx 569\;\mathrm{K} \approx 296\;^\circ\mathrm{C}}$

Despite the $8$ litres of Fanta, the rock wins out: its heat capacity ($1200\;\mathrm{J/K}$) is more than three times that of the Fanta ($360\;\mathrm{J/K}$), because the rock’s specific heat is ten times greater. The equilibrium therefore remains scalding hot.

Collegamenti

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