Solved thermodynamics exercise

From cold ice to boiling water: orders of magnitude

Exercise 1 · Lesson 2 — A History of Thermodynamics and Calorimetry

calorimetry
phase change
melting
vaporization
latent heat
order of magnitude

Statement

Numerical data (at atmospheric pressure):

\begin{aligned} c_{\text{ice}} &= 2{,}1\times10^3\,\text{J}\!\cdot\!\text{kg}^{-1}\!\cdot\!\text{K}^{-1}, & L_f &= 3{,}34\times10^5\,\text{J}\!\cdot\!\text{kg}^{-1},\\ c_{\text{water}} &= 4{,}18\times10^3\,\text{J}\!\cdot\!\text{kg}^{-1}\!\cdot\!\text{K}^{-1}, & L_v &= 2{,}26\times10^6\,\text{J}\!\cdot\!\text{kg}^{-1}. \end{aligned}

At atmospheric pressure, one kilogram of ice, initially at $-100\,{}^\circ\text{C}$ , is heated progressively.

Calculate the energy required to heat the ice from $-100\,{}^\circ\text{C}$ to $0\,{}^\circ\text{C}$ .
Calculate the energy required to melt all the ice at $0\,{}^\circ\text{C}$ .
Calculate the energy required to heat the liquid water from $0\,{}^\circ\text{C}$ to $100\,{}^\circ\text{C}$ .
Calculate the additional energy required to vaporize all the water at $100\,{}^\circ\text{C}$ .
Which stage requires the most energy? Compare the energy needed to heat one kilogram of water from $0\,{}^\circ\text{C}$ to $100\,{}^\circ\text{C}$ with the gravitational potential energy of a mass $m$ falling through ten metres. What mass must fall to release the same energy?
Now consider the reverse process. A mass $m_v$ of water vapour condenses on a windscreen, without the resulting liquid water subsequently cooling. Express the heat $Q_{\text{rel}}$ released during condensation, then calculate it for $m_v=1{,}0\times10^{-2}\,\text{kg}$ . At the windscreen temperature, take $L_v=2{,}45\times10^6\,\text{J}\!\cdot\!\text{kg}^{-1}$ .

Hint

To heat a phase without a phase change, use

Q=mc\Delta T

. For a phase change at constant temperature, use

Q=mL

. The gravitational potential energy of a mass

m

at height

h

E_p=mgh

; take

g=9{,}81\,\text{m}\!\cdot\!\text{s}^{-2}

Detailed solution

Solution

1. Heating the ice to its melting temperature requires

Q_1=mc_{\text{ice}}\Delta T =1{,}0\times2{,}1\times10^3\times100 =2{,}1\times10^5\,\text{J}.

2. During melting, the temperature remains at $0\,{}^\circ\text{C}$ :

Q_2=mL_f=1{,}0\times3{,}34\times10^5 =3{,}34\times10^5\,\text{J}.

3. To heat the liquid water from $0$ to $100\,{}^\circ\text{C}$ ,

Q_3=mc_{\text{water}}\Delta T =1{,}0\times4{,}18\times10^3\times100 =4{,}18\times10^5\,\text{J}.

4. Complete vaporization requires a further

Q_4=mL_v=1{,}0\times2{,}26\times10^6 =2{,}26\times10^6\,\text{J}.

This stage alone therefore requires a few megajoules.

5. Vaporization is by far the most energy-intensive stage:

Q_4=2{,}26\times10^6\,\text{J} >Q_3=4{,}18\times10^5\,\text{J} >Q_2=3{,}34\times10^5\,\text{J} >Q_1=2{,}1\times10^5\,\text{J}.

In particular, vaporization requires about $(2{,}26\times10^6)/(4{,}18\times10^5)\approx5{,}4$ times as much energy as heating the liquid water from $0$ to $100\,{}^\circ\text{C}$ .

For a mass $m$ falling through $h=10\,\text{m}$ , $E_p=mgh$ . Setting $mgh=Q_3$ gives

m=\frac{Q_3}{gh} =\frac{4{,}18\times10^5}{9{,}81\times10} \approx4{,}26\times10^3\,\text{kg}.

A mass of about $4{,}3$ tonnes would therefore have to fall ten metres to release as much energy as is needed to heat one kilogram of water from $0$ to $100\,{}^\circ\text{C}$ ! This is enormous and explains why water heating accounts for a significant share of domestic energy use. Water also has a very large specific heat capacity, especially compared with common metals: it is about ten times that of copper, for example (see the following exercises).

6. Condensation is the reverse of vaporization: the vapour releases the latent heat

Q_{\text{rel}}=m_vL_v.

For $m_v=1{,}0\times10^{-2}\,\text{kg}$ ,

Q_{\text{rel}}=1{,}0\times10^{-2}\times2{,}45\times10^6 =2{,}45\times10^4\,\text{J}.

Thus, even a small mass of condensing vapour can release a non-negligible amount of heat, received by the windscreen and its surroundings.