IPhO 1969 · Theory problem 2

Ice dropped into water in a copper calorimeter

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Water of mass \(m_2\) is contained in a copper calorimeter of mass \(m_1\). Their common temperature is \(t_2\). A piece of ice of mass \(m_3\) and temperature \(t_3<0^\circ\mathrm{C}\) is dropped into the calorimeter.

a) Determine the temperature and masses of water and ice in the equilibrium state for general values of \(m_1,m_2,m_3,t_2\) and \(t_3\). Write equilibrium equations for all possible processes which have to be considered.

b) Find the final temperature and final masses of water and ice for \(m_1=1.00\ \mathrm{kg}\), \(m_2=1.00\ \mathrm{kg}\), \(m_3=2.00\ \mathrm{kg}\), \(t_2=10^\circ\mathrm{C}\), \(t_3=-20^\circ\mathrm{C}\).

Neglect the energy losses, assume the normal barometric pressure. Specific heat of copper is \(c_1=0.1\ \mathrm{kcal/kg\cdot{}^\circ C}\), specific heat of water \(c_2=1\ \mathrm{kcal/kg\cdot{}^\circ C}\), specific heat of ice \(c_3=0.492\ \mathrm{kcal/kg\cdot{}^\circ C}\), latent heat of fusion of ice \(l=78,7\ \mathrm{kcal/kg}\). Take \(1\ \mathrm{cal}=4.2\ \mathrm{J}\).

Official English text of the problem.

Answers

  • a) Define the total water-plus-ice mass and the heat surplus at \(0^\circ\mathrm{C}\): \[ M=m_2+m_3,\qquad D=(m_1c_1+m_2c_2)t_2+m_3c_3t_3. \] The equilibrium temperature \(t_f\) and masses are:
Condition \(t_f\) Final liquid-water mass Final ice mass
\(D<-m_2l\) \(\displaystyle\frac{D+m_2l}{m_1c_1+Mc_3}<0\) \(0\) \(M\)
\(-m_2l\le D\le m_3l\) \(0^\circ\mathrm{C}\) \(\displaystyle m_2+\frac{D}{l}\) \(\displaystyle m_3-\frac{D}{l}\)
\(D>m_3l\) \(\displaystyle\frac{D-m_3l}{m_1c_1+Mc_2}>0\) \(M\) \(0\)

In the middle regime, \(D<0\) causes freezing, \(D>0\) causes melting, and \(D=0\) leaves the phase masses unchanged. The endpoints correspond to all ice or all water at \(0^\circ\mathrm{C}\).

  • b) \[ \boxed{t_f=0^\circ\mathrm{C},\qquad m_{\mathrm{water}}=0.890\ \mathrm{kg},\qquad m_{\mathrm{ice}}=2.11\ \mathrm{kg}.} \] Approximately \(0.110\ \mathrm{kg}\) of the original water freezes.

Solution

a)

With no heat loss, the heat released by cooling and freezing equals the heat absorbed by warming and melting. We use the supplied specific heats and latent heat as constants. At normal barometric pressure, liquid water and ice coexist in equilibrium at \(0^\circ\mathrm{C}\).

Let \[ A=(m_1c_1+m_2c_2)t_2 \] be the heat released when the calorimeter and the original water cool to \(0^\circ\mathrm{C}\), and let \[ B=m_3c_3(0-t_3) \] be the heat required to warm the original ice to \(0^\circ\mathrm{C}\). Their difference is \[ D=A-B=(m_1c_1+m_2c_2)t_2+m_3c_3t_3. \]

The possible processes follow from comparing this difference with the latent heats available.

Partial melting, with final temperature \(0^\circ\mathrm{C}\).

If \(0\le D\le m_3l\), the sensible heat surplus melts a mass \(x\) of ice. The equilibrium equation is \[ (m_1c_1+m_2c_2)t_2=m_3c_3(0-t_3)+xl, \] or \(A=B+xl\). Thus \[ x=\frac{D}{l},\qquad t_f=0^\circ\mathrm{C}, \] \[ m_{\mathrm{water}}=m_2+\frac{D}{l}, \qquad m_{\mathrm{ice}}=m_3-\frac{D}{l}. \] At \(D=0\), no net phase change occurs. At \(D=m_3l\), the last ice melts and the final state is entirely water at \(0^\circ\mathrm{C}\).

Complete melting, with final temperature above \(0^\circ\mathrm{C}\).

If \(D>m_3l\), all the ice melts. The calorimeter and original water cool from \(t_2\) to \(t_f\), while the added ice warms to zero, melts, and then warms as liquid water to \(t_f\): \[ (m_1c_1+m_2c_2)(t_2-t_f) = m_3c_3(0-t_3)+m_3l+m_3c_2t_f. \] Collecting the terms containing \(t_f\) gives \[ D-m_3l=\bigl[m_1c_1+(m_2+m_3)c_2\bigr]t_f. \] Therefore \[ t_f=\frac{D-m_3l}{m_1c_1+(m_2+m_3)c_2}, \qquad m_{\mathrm{water}}=m_2+m_3,\qquad m_{\mathrm{ice}}=0. \]

Partial freezing, with final temperature \(0^\circ\mathrm{C}\).

If \(-m_2l\le D<0\), cooling the calorimeter and water to zero supplies insufficient heat to warm the original ice to zero. A mass \(y\) of the original water must freeze, releasing latent heat: \[ (m_1c_1+m_2c_2)t_2+yl=m_3c_3(0-t_3). \] Hence \[ y=-\frac{D}{l},\qquad t_f=0^\circ\mathrm{C}, \] \[ m_{\mathrm{water}}=m_2-y=m_2+\frac{D}{l}, \qquad m_{\mathrm{ice}}=m_3+y=m_3-\frac{D}{l}. \] At \(D=-m_2l\), all the original water freezes, leaving only ice at \(0^\circ\mathrm{C}\).

Complete freezing, with final temperature below \(0^\circ\mathrm{C}\).

If \(D<-m_2l\), even freezing all the original water cannot bring the system to zero. The final state contains only ice at \(t_f<0\).

The calorimeter cools from \(t_2\) to \(t_f\). The original water cools to zero, freezes, and then cools as ice to \(t_f\). Their released heat warms the original ice from \(t_3\) to \(t_f\): \[ m_1c_1(t_2-t_f) +m_2c_2t_2 +m_2l +m_2c_3(0-t_f) = m_3c_3(t_f-t_3). \] Rearranging, \[ D+m_2l=\bigl[m_1c_1+(m_2+m_3)c_3\bigr]t_f, \] so \[ t_f=\frac{D+m_2l}{m_1c_1+(m_2+m_3)c_3}, \qquad m_{\mathrm{water}}=0,\qquad m_{\mathrm{ice}}=m_2+m_3. \]

These cases exhaust the possible equilibrium states.

b)

The calorimeter and water can release \[ A=(1.00\times0.1+1.00\times1)\times10 =11.0\ \mathrm{kcal} \] by cooling to \(0^\circ\mathrm{C}\).

The heat needed to warm the original ice to zero is \[ B=2.00\times0.492\times20 =19.68\ \mathrm{kcal}. \] Consequently, \[ D=A-B=-8.68\ \mathrm{kcal}. \] Since \[ -m_2l=-78.7\ \mathrm{kcal} < D < 0, \] only part of the original water freezes, and the equilibrium temperature is \(0^\circ\mathrm{C}\).

The mass that freezes is \[ y=\frac{B-A}{l} =\frac{8.68\ \mathrm{kcal}}{78.7\ \mathrm{kcal/kg}} =0.110292\ldots\ \mathrm{kg}. \] Thus \[ \boxed{m_{\mathrm{water}}=1.00-y=0.890\ \mathrm{kg}}, \] \[ \boxed{m_{\mathrm{ice}}=2.00+y=2.11\ \mathrm{kg}}, \qquad \boxed{t_f=0^\circ\mathrm{C}}. \]

The solution was checked against the official solution.