Solved thermodynamics exercise

Determining the specific heat capacity of a metal by the method of mixtures

Exercise 4 · Lesson 2 — A History of Thermodynamics and Calorimetry

  • calorimetry
  • method of mixtures
  • specific heat capacity
  • calorimeter
  • water equivalent

Statement

As Joseph Black did in the eighteenth century, we wish to measure the specific heat capacity cmc_m of an unknown metal by the method of mixtures. A sample of mass m=0,150 kgm=0{,}150\,\text{kg} is heated to Tm=95 ∘CT_m=95\,^\circ\text{C} and quickly immersed in a calorimeter containing me=0,250 kgm_e=0{,}250\,\text{kg} of water, with ce=4,18×103 J ⁣⋅ ⁣kg−1 ⁣⋅ ⁣K−1c_e=4{,}18\times10^3\,\text{J}\!\cdot\!\text{kg}^{-1}\!\cdot\!\text{K}^{-1}, initially at Te=18 ∘CT_e=18\,^\circ\text{C}. The measured equilibrium temperature is Teq=22 ∘CT_{eq}=22\,^\circ\text{C}. Initially, neglect the heat capacity of the calorimeter vessel, stirrer, and thermometer.

  1. Write the energy-conservation equation for the isolated system consisting of the metal and water, with cmc_m as the only unknown.
  2. Deduce an expression for cmc_m and calculate it. Which common metal does it approximately match? Reference values, in J ⁣⋅ ⁣kg−1 ⁣⋅ ⁣K−1\text{J}\!\cdot\!\text{kg}^{-1}\!\cdot\!\text{K}^{-1}, are: aluminium 9,0×1029{,}0\times10^2, iron 4,5×1024{,}5\times10^2, copper 3,9×1023{,}9\times10^2, and lead 1,3×1021{,}3\times10^2.
  3. The calorimeter vessel actually absorbs some heat. Model this by a water equivalent μ=1,5×10−2 kg\mu=1{,}5\times10^{-2}\,\text{kg}, as though the calorimeter were an additional mass μ\mu of water. Recalculate cmc_m and comment on the direction of the correction.
  4. Why must the sample be immersed quickly and the calorimeter be well insulated from the surrounding air?

Hint
The metal releases Qm=mcm(Tm−Teq)Q_m=mc_m(T_m-T_{eq}), received by the water and, in Question 3, by the calorimeter: Qm=(me+μ)ce(Teq−Te)Q_m=(m_e+\mu)c_e(T_{eq}-T_e).

Detailed solution

Solution
1. Energy conservation gives

mcm(Tm−Teq)=mece(Teq−Te).mc_m(T_m-T_{eq})=m_ec_e(T_{eq}-T_e).

2. Thus

cm=mece(Teq−Te)m(Tm−Teq)=0,250×4,18×103×(22−18)0,150×(95−22)≈3,82×102 J ⁣⋅ ⁣kg−1 ⁣⋅ ⁣K−1.c_m=\frac{m_ec_e(T_{eq}-T_e)}{m(T_m-T_{eq})} =\frac{0{,}250\times4{,}18\times10^3\times(22-18)}{0{,}150\times(95-22)} \approx3{,}82\times10^2\,\text{J}\!\cdot\!\text{kg}^{-1}\!\cdot\!\text{K}^{-1}.

This is very close to the value for copper.

3. Including the calorimeter's water equivalent,

cm=(me+μ)ce(Teq−Te)m(Tm−Teq)=(0,250+0,015)×4,18×103×40,150×73≈4,05×102 J ⁣⋅ ⁣kg−1 ⁣⋅ ⁣K−1.c_m=\frac{(m_e+\mu)c_e(T_{eq}-T_e)}{m(T_m-T_{eq})} =\frac{(0{,}250+0{,}015)\times4{,}18\times10^3\times4}{0{,}150\times73} \approx4{,}05\times10^2\,\text{J}\!\cdot\!\text{kg}^{-1}\!\cdot\!\text{K}^{-1}.

The correction increases cmc_m: neglecting the vessel underestimates the heat actually released by the metal and therefore underestimates its specific heat capacity.

4. The method assumes an isolated system throughout the measurement. Rapid immersion limits the time during which the hot sample can exchange heat with the surrounding air before reaching the water, while good insulation limits losses to the surroundings as equilibrium is established. Any unaccounted heat transfer would distort the balance in Question 1 and hence the inferred value of cmc_m. This is precisely the kind of experimental care already taken by Black, and later by Lavoisier and Laplace with their ice calorimeter.