Aluminium block in nitrogen (thermal equilibrium)

Problem

An aluminium block of mass $m_\text{Al} = 0{,}5\ \text{kg}$ ($c_s = 900\ \text{J/(kg·K)}$) at $T = 400\ \text{K}$ is placed in an insulated rigid cylinder containing $n = 1\ \text{mol}$ of nitrogen (diatomic, $f = 5$) at $T = 290\ \text{K}$. Find the equilibrium temperature (neglect the volume of the block).

Solution

Setup. The container is insulated and rigid: no heat is exchanged with the surroundings and no work is done. The total internal energy is conserved, i.e. the sum of the changes is zero: $m_\text{Al}\, c_s\,(T_f - 400) + \frac{5}{2}\, n R\,(T_f - 290) = 0$

Coefficients. $m_\text{Al}\, c_s = 0{,}5\cdot 900 = 450\ \text{J/K}, \qquad \frac{5}{2}\, n R = \frac{5}{2}\cdot 1\cdot 8{,}31 = 20{,}775\ \text{J/K}$

Equation. $450\,(T_f - 400) + 20{,}775\,(T_f - 290) = 0$ $470{,}775\, T_f = 180000 + 6024{,}75 = 186024{,}75$ $T_f = \frac{186024{,}75}{470{,}775} = 395{,}2\ \text{K}$

$\ev{T_f \approx 395\ \text{K}}$

The process is irreversible (heat exchange by contact between bodies at different temperatures), but energy is conserved regardless.

Links

Topics: Thermodynamics Concepts: First law of thermodynamics · Temperature Skills: Calorimetric equation Objects: Ideal gas · Calorimeter