Temperature — Question 1

Question 1

Dividing an equilibrium body does not divide its temperature

A uniform equilibrium block has mass $6.0\,\mathrm{kg}$ and temperature $320\,\mathrm K$. It is separated without energy exchange or deformation into masses $1.5\,\mathrm{kg}$ and $4.5\,\mathrm{kg}$. Use the illustrative model $U=mcT$ with constant $c=500\,\mathrm{J\,kg^{-1}K^{-1}}$ and a common energy reference.

1. Find the temperatures and modeled internal energies of the two pieces.

2. Rejoin the pieces under the same ideal conditions. Explain quantitatively why temperature is intensive although internal energy is additive.

Question 1 - Solution

Dividing an equilibrium body does not divide its temperature

(a) Method and calculation

Each piece inherits the same local equilibrium state:$$T_1=T_2=320\,\mathrm K.$$$$U_1=1.5(500)(320)=240000\,\mathrm J,\quad U_2=4.5(500)(320)=720000\,\mathrm J.$$Division changes the quantity of material, not its temperature.

(b) Interpretation and further calculation

Energy and mass add:$$U=U_1+U_2=960000\,\mathrm J,\quad m=m_1+m_2=6.0\,\mathrm{kg}.$$$$T=\frac{U}{mc}=\frac{960000}{6(500)}=320\,\mathrm K.$$Adding the two temperature readings to obtain $640\,\mathrm K$ has no physical basis.

Physical check

The energy model is supplied for this exercise; a temperature value alone does not determine energy without mass, material and reference information.