Solved thermodynamics exercise

Mixing two masses of water

Exercise 3 · Lesson 2 — A History of Thermodynamics and Calorimetry

  • calorimetry
  • mixing
  • heat capacity
  • specific heat capacity
  • Joseph Black
  • equilibrium temperature

Statement

A mass m1=0,200 kgm_1=0{,}200\,\text{kg} of water at T1=80 ∘CT_1=80\,^\circ\text{C} is poured into a vessel already containing m2=0,300 kgm_2=0{,}300\,\text{kg} of water at T2=15 ∘CT_2=15\,^\circ\text{C}. The vessel is an ideal calorimeter: heat loss to the surroundings and the heat capacity of the vessel itself are neglected. Recall that Q=mcΔTQ=mc\Delta T, where c=4,18×103 J ⁣⋅ ⁣kg−1 ⁣⋅ ⁣K−1c=4{,}18\times10^3\,\text{J}\!\cdot\!\text{kg}^{-1}\!\cdot\!\text{K}^{-1} is the specific heat capacity of liquid water.

  1. Justify that the heat released by the hot water is entirely received by the cold water. Write the conservation equation involving m1m_1, m2m_2, cc, T1T_1, T2T_2, and the equilibrium temperature TeqT_{eq}.
  2. Deduce an expression for TeqT_{eq} and calculate its numerical value.
  3. Calculate the heat QQ released by the hot water during mixing.
  4. Can mixtures of the same substance be used to determine cc? Explain.

Hint
Because the calorimeter is isolated and rigid, the total internal energy U1+U2U_1+U_2 is conserved: the heat released by one mass is received by the other, with the opposite sign.

Detailed solution

Solution
1. Conservation of the total internal energy of the system gives

m1c(Teq−T1)+m2c(Teq−T2)=0.m_1c(T_{eq}-T_1)+m_2c(T_{eq}-T_2)=0.

2. The common coefficient cc cancels, giving

Teq=m1T1+m2T2m1+m2=0,200×80+0,300×150,200+0,300=41 ∘C.T_{eq}=\frac{m_1T_1+m_2T_2}{m_1+m_2} =\frac{0{,}200\times80+0{,}300\times15}{0{,}200+0{,}300} =41\,^\circ\text{C}.

3. The heat released by the hot water is

Q=m1c(T1−Teq)=0,200×4,18×103×(80−41)≈3,26×104 J.Q=m_1c(T_1-T_{eq}) =0{,}200\times4{,}18\times10^3\times(80-41) \approx3{,}26\times10^4\,\text{J}.

The cold water receives the same quantity:

m2c(Teq−T2)=0,300×4,18×103×26≈3,26×104 J.m_2c(T_{eq}-T_2) =0{,}300\times4{,}18\times10^3\times26 \approx3{,}26\times10^4\,\text{J}.

4. No. For two masses of the same substance, cc multiplies both terms and cancels. The measured equilibrium temperature is therefore independent of cc and is simply the mass-weighted mean of the initial temperatures. Measuring a specific heat capacity by the method of mixtures requires another substance whose heat capacity is known, as in the next exercise.