Heating a Block of Ice!

Assume that the ice gets to each state and find the mass or the temperature accordingly to see if the assumption was correct.

First, let’s write the conservation of energy for this specific calorimetry problem. It reads:

\begin{equation*}
Q_{\text{in}}=\sum_{j=1}^N Q_j.
\end{equation*}

The heat for temperature change is:

\begin{equation*}
Q=mc\Delta T,
\end{equation*}

and the latent heat, for change of phase in matter, is:

\begin{equation}
Q= \pm mL.
\end{equation}

Let \(Q_1\) be the heat to raise the temperature of the ice to \(0 ^\circ \text{C} \). \(Q_2\) the latent heat for fusion. \(Q_3\) the heat as water to raise the temperature from \(0 ^\circ \text{C} \) to \(100 ^\circ \text{C} \). \(Q_4\) the latent heat of vaporization. Finally, \(Q_5\) the heat as vapor to raise the temperature from \(100 ^\circ \text{C} \) to a final temperature \(T_f\). The total heat that enters to the system is:

\begin{equation*}
Q_{\text{in}}=Q_1+Q_2+Q_3+Q_4+Q_5,
\end{equation*}

or:

\begin{equation*}
Q_{\text{in}}=mc_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})+mL_{\text{f}}+mc_{\text{water}}(T_f^{\text{water}}-T_i^{\text{water}})\end{equation*} \begin{equation*}\label{sup5.1}+mL_{\text{v}}+mc_{\text{vapor}}(T_{f}^{\text{vapor}}-T_i^{\text{vapor}}).
\end{equation*}

Solving for \(T_f\) we get:

\begin{equation*}
T_{f}^{\text{vapor}}=T_i^{\text{vapor}}+\frac{Q_{\text{in}}}{mc_{\text{vapor}}}-\frac{c_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})}{c_{\text{vapor}}}-\frac{L_{\text{f}}}{c_{\text{vapor}}}\end{equation*} \begin{equation*}-\frac{c_{\text{water}}(T_f^{\text{water}}-T_i^{\text{water}})}{c_{\text{vapor}}}-\frac{L_{\text{v}}}{c_{\text{vapor}}}.
\end{equation*}

With numerical values:

\begin{equation*}
T_f^{\text{vapor}}\approx 568\,{}^{\circ}\text{C}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

They’ve asked us to find the final temperature and phase of the ice. To solve this calorimetry problem, we must use the conservation of energy equation applied to this particular problem and several assumptions about the final state of the ice. Then, by finding the actual final state of the ice, we’ll verify if our assumptions are correct.

First, let’s write the conservation of energy for this specific calorimetry problem. It reads

\begin{equation}
\label{sumaQ}
Q_{\text{in}}=\sum_{j=1}^N Q_j,
\end{equation}

where \(Q_{\text{in}}\) is the amount of heat that is added to the system and \(Q_j\) are the different types of heat absorbed by the ice. The explicit expression of each \(Q_j\) depends on the process and the state of matter in which the system is. The total number of distinct processes that occur is denoted by \(N\). The system in our case is the \(3\,\text{kg}\) of ice.

There are two types of forms that the terms \(Q_j\) can take:

(i) Energy is used for temperature change, namely

\begin{equation}
\label{mcdt}
Q=mc\Delta T,
\end{equation}

where \(m\) is the mass of the object, \(c\) its specific heat, and \(\Delta T=T_f-T_i\) the change from an initial temperature \(T_i\) to a final one \(T_f\).

(ii) Energy is used for a change of phase in the matter. This occurs at a constant temperature, and the expression for \(Q\) will be

\begin{equation}
\label{mL}
Q=\pm mL,
\end{equation}

where \(m\) is the amount of mass of the object that makes the phase transition and \(L\) is the latent heat, which is different for different elements and different phase transitions. The positive sign is chosen when the object transitions from solid to liquid or liquid to gas, and the negative sign is chosen when the object transitions from liquid to solid or gas to liquid.

Now, let’s start with the assumptions. For our first assumption, we assume that the final state of the system is entirely ice with a temperature \(T_f<0\,{}^{\circ}\text{C}\), so no change of phase occurs and only one process takes place \(N=1\), namely the change of temperature of ice. Thus, we write equation \eqref{sumaQ} as

\begin{equation}
\label{sup1}
Q_{\text{in}}=Q_1.
\end{equation}

We must the use equation \eqref{mcdt} to calculate \(Q_1\) in the corresponding phase; explicitly,

\begin{equation}
\label{q1}
Q_1=m_{\text{ice}}c_{\text{ice}}(T_f-T_i).
\end{equation}

Using the expression given in equation \eqref{q1} into equation \eqref{sup1}, we obtain

\begin{equation}
Q_{\text{in}}=m_{\text{ice}}c_{\text{ice}}(T_f-T_i),
\end{equation}

where we can divide both sides by \(m_{\text{ice}}c_{\text{ice}}\) to get

\begin{equation}
\frac{Q_{\text{in}}}{m_{\text{ice}}c_{\text{ice}}}=T_f-T_i,
\end{equation}

and solve for \(T_f\) to obtain

\begin{equation}
T_f=T_i+\frac{Q_{\text{in}}}{m_{\text{ice}}c_{\text{ice}}}.
\end{equation}

Calculating \(T_f\) explicitly with the given numerical values, we obtain

\begin{equation}
T_f=-20\,{}^{\circ}\text{C}+\frac{1.2\times10^7\,\text{J}}{(3\,\text{kg})(2108\,\text{J/kg\,}{}^{\circ}\text{C})}\approx1877\,{}^{\circ}\text{C}.
\end{equation}

Clearly, this temperature makes no sense because for our first assumption, we said \(T_f<0\,{}^{\circ}\text{C}\) in order to have only ice. Thus, the first assumption turned out to be wrong.

We’ll follow with the second assumption: the ice melts partially, thus creating a mixed phase of ice and water. The final temperature is the only temperature at which this mixed-phase can exist at equilibrium, which is the temperature of the phase transition \(T_f=0\,{}^{\circ}\text{C}\). Notice that now we have two processes: the change of temperature of ice and the phase transition of ice into water (solid to liquid); therefore, \(N=2\). The term \(Q_1\) is taken as in \eqref{q1}. The term \(Q_2\) must be calculated following equation \eqref{mL} with the appropriate mass and latent heat for fusion, namely

\begin{equation}
\label{q2}
Q_2=m_{\text{ice melts}}L_{\text{f}},
\end{equation}

where \(L_{\text{f}}\) is the latent heat of fusion and \(m_{\text{ice melts}}\) is the amount of ice that melts. For our second assumption to make sense, this amount of mass of ice that melts must be less or equal to the total mass of ice, that is \(m_{\text{ice melts}}\leq 3\,\text{kg}\). Using equation \eqref{sumaQ} for our second assumption, we obtain

\begin{equation}
\label{sup2}
Q_{\text{in}}=Q_1+Q_2.
\end{equation}

Using the expressions of equations \eqref{q1} and \eqref{q2} into equation \eqref{sup2}, we obtain

\begin{equation}
\label{sup2.1}
Q_{\text{in}}=m_{\text{ice}}c_{\text{ice}}(T_f-T_i)+m_{\text{ice melts}}L_{\text{f}},
\end{equation}

where our unknown is now \(m_{\text{ice melts}}\). Solving for \(m_{\text{ice melts}}\) in equation \eqref{sup2.1}, we get

\begin{equation}
m_{\text{ice melts}}=\frac{Q_{\text{in}}}{L_{\text{f}}}-\frac{m_{\text{ice}}c_{\text{ice}}(T_f-T_i)}{L_{\text{f}}}.
\end{equation}

Using the numerical values for our second assumption (\(T_f=0\,{}^{\circ}\text{C}\)), we obtain

\begin{equation}
m_{\text{ice melts}}=\frac{1.2\times 10^7\,\text{J}}{334000\,\text{J/kg}}-\frac{(3\,\text{kg})(2108\,\text{J/kg\,}{}^{\circ}\text{C})(0\,{}^{\circ}\text{C}-(-20\,{}^{\circ}\text{C}))}{334000\,\text{J/kg}},
\end{equation}

\begin{equation}
m_{\text{ice melts}}\approx 36\,\text{kg}.
\end{equation}

From the result above, it is clear that our second assumption is also incorrect since \(m_{\text{ice melts}}\) must be less than or equal to \(3\,\text{kg}\). So, we continue with a third assumption.

In our third assumption, we’ll consider that all the ice is melted and the resulting water has a final temperature \(T_f\), which we must find, between the values \(0\,{}^{\circ}\text{C}<T_f<100\,{}^{\circ}\text{C}\). We must then add a term concerning the heating of the water that once was ice, then \(N=3\) and equation \eqref{sumaQ} becomes \begin{equation} \label{sup3} Q_{\text{in}}=Q_1+Q_2+Q_3. \end{equation} The term \(Q_3\) is the one concerning the water’s rise in temperature, and it can be written accordingly to equation \eqref{mcdt} as \begin{equation} \label{q3} Q_3=m_{\text{water}}c_{\text{water}}(T_f^{\text{water}}-T_i^{\text{water}}), \end{equation} where we have made the distinction in the temperatures, mass, and specific heat to indicate that these variables correspond to water. The terms \(Q_1\) and \(Q_2\) can be obtained from equations \eqref{q1} and \eqref{q2}, only in this case \(m_{\text{ice melts}}=m_{\text{ice}}=m_{\text{water}}\) because no mass is lost during the phase transition. We can then rewrite equation \eqref{sup3} as \begin{equation} \label{sup3.1} Q_{\text{in}}=m_{\text{ice}}c_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})+m_{\text{ice}}L_{\text{f}}+m_{\text{water}}c_{\text{water}}(T_f^{\text{water}}-T_i^{\text{water}}), \end{equation} where we have also made the distinction on the temperatures at which ice begins \(T_i^{\text{ice}}=-20\,{}^{\circ}\text{C}\) and at which ice becomes water \(T_f^{\text{ice}}=0\,{}^{\circ}\text{C}\). As a consequence, the numerical value for \(T_i^{\text{water}}=0\,{}^{\circ}\text{C}\). Solving for \(T_f^{\text{water}}\), we have \begin{equation} Q_{\text{in}}-m_{\text{ice}}c_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})-m_{\text{ice}}L_{\text{f}}=m_{\text{water}}c_{\text{water}}(T_f^{\text{water}}-T_i^{\text{water}}), \end{equation}, and dividing both sides by \(m_{\text{water}}c_{\text{water}}\) \begin{equation} \frac{Q_{\text{in}}}{m_{\text{water}}c_{\text{water}}}-\frac{m_{\text{ice}}c_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})}{m_{\text{water}}c_{\text{water}}}-\frac{m_{\text{ice}}L_{\text{f}}}{m_{\text{water}}c_{\text{water}}}=T_f^{\text{water}}-T_i^{\text{water}}, \end{equation} where we can solve directly for \(T_f^{\text{water}}\) to get \begin{equation} \label{sup3.2} T_f^{\text{water}}=T_i^{\text{water}}+\frac{Q_{\text{in}}}{m_{\text{water}}c_{\text{water}}}-\frac{m_{\text{ice}}c_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})}{m_{\text{water}}c_{\text{water}}}-\frac{m_{\text{ice}}L_{\text{f}}}{m_{\text{water}}c_{\text{water}}}. \end{equation} Notice that since \(m_{\text{water}}=m_{\text{ice}}\) we can further simplify the expression on equation \eqref{sup3.2} to \begin{equation} T_f^{\text{water}}=T_i^{\text{water}}+\frac{Q_{\text{in}}}{m_{\text{water}}c_{\text{water}}}-\frac{c_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})}{c_{\text{water}}}-\frac{L_{\text{f}}}{c_{\text{water}}}. \end{equation} Using the numerical values we obtain \begin{equation} T_f^{\text{water}}=\frac{1.2\times 10^7\,\text{J}}{(3\,\text{kg})(4186\,\text{J/kg}\,{}^{\circ}\text{C})}-\frac{(2108\,\text{J/kg}\,{}^{\circ}\text{C})(-(-20\,{}^{\circ}\text{C}))}{4186\,\text{J/kg}\,{}^{\circ}\text{C}}-\frac{334000\,\text{J/kg}}{4186\,\text{J/kg}\,{}^{\circ}\text{C}}, \end{equation} \begin{equation} T_f^{\text{water}}\approx 885.8\,{}^{\circ}\text{C}. \end{equation} Once more, the temperature is too high and the third assumption is not correct because \(T_f>100\,{}^{\circ}\text{C}\). Now it’s time to make a fourth assumption.

In this fourth assumption, the ice rises in temperature, then it melts, then the water rises in temperature and then it vaporizes partially such that in the final state we have water and vapor coexisting. The final temperature of the system would then be \(T_f=100\,{}^{\circ}\text{C}\), the temperature of the phase transition between liquid and gas. As before, we identify four process (\(N=4\)), so equation \eqref{sumaQ} becomes

\begin{equation}
\label{sup4}
Q_{\text{in}}=Q_1+Q_2+Q_3+Q_4.
\end{equation}

This time, \(Q_4\) will take the form given by equation \eqref{mL} with the appropriate labels for the corresponding phase transition, that is

\begin{equation}
\label{q4}
Q_4=m_{\text{water boils}}L_{\text{v}},
\end{equation}

where \(L_{\text{v}}\) is the latent heat of vaporization and \(m_{\text{water boils}}\) is the amount of water that boils and becomes vapor; clearly this quantity cannot be greater than the initial mass of ice, that is \(m_{\text{water boils}}\leq 3\,\text{kg}\). The terms \(Q_1\,Q_2\) and \(Q_3\) are taken as in equations \eqref{q1}, \eqref{q2} and \eqref{q3} respectively. Putting everything together into equation \eqref{sup4}, we finally get \begin{equation} Q_{\text{in}}=m_{\text{ice}}c_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})+m_{\text{ice}}L_{\text{f}}+m_{\text{water}}c_{\text{water}}(T_f^{\text{water}}-T_i^{\text{water}})+m_{\text{water boils}}L_{\text{v}}. \end{equation} Solving for \(m_{\text{water boils}}\) we obtain \begin{equation} m_{\text{water boils}}=\frac{Q_{\text{in}}}{L_\text{v}}-\frac{m_{\text{ice}}c_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})}{L_\text{v}}-\frac{m_{\text{ice}}L_{\text{f}}}{L_\text{v}}-\frac{m_{\text{water}}c_{\text{water}}(T_f^{\text{water}}-T_i^{\text{water}})}{L_\text{v}}. \end{equation} Using the numerical values, we get \begin{equation*} m_{\text{water boils}}= \frac{1.2\times 10^7\,\text{J}}{2.27\times 10^{6}\,\text{J/kg}}-\frac{(3\,\text{kg})(2108\,\text{J/kg}\,{}^{\circ}\text{C})(-(-20\,{}^{\circ}\text{C}))}{2.27\times 10^6\,\text{J/kg}} \end{equation*} \begin{equation} -\frac{(3\,\text{kg})(334000\,\text{J/kg})}{2.27\times 10^6\,\text{J/kg}}-\frac{(3\,\text{kg})(4186\,\text{J/kg}\,{}^{\circ}\text{C})(100\,{}^{\circ}\text{C})}{2.27\times 10^6\,\text{J/kg}}, \end{equation} \begin{equation} m_{\text{water boils}}\approx 4.89\,\text{kg}. \end{equation} Clearly, assumption 4 is wrong too because \(m_{\text{water boils}}>3\,\text{kg}\). We are only left with assumption five.

In assumption five, the ice rises in temperature, then it melts into water completely. This water then rises in temperature and evaporates completely. Finally, the resulting vapor rises its temperature to a final temperature \(T_f>100\,{}^{\circ}\text{C}\). In total, we have five processes, so \(N=5\) and equation \eqref{sumaQ} can be written as

\begin{equation}
\label{sup5}
Q_{\text{in}}=Q_1+Q_2+Q_3+Q_4+Q_5,
\end{equation}

where \(Q_1,\,Q_2,\,Q_3\) and \(Q_4\) are given by equations \eqref{q1}, \eqref{q2}, \eqref{q3} and \eqref{q4} respectively. The equation for \(Q_5\) corresponds to the vapor rising its temperature. We must use the form of equation \eqref{mcdt} with the appropriate labels, namely,

\begin{equation}
Q_5=m_{\text{vapor}}c_{\text{vapor}}(T_{f}^{\text{vapor}}-T_i^{\text{vapor}}).
\end{equation}

The initial temperature of vapor is clearly that of the phase transition, that is \(T_{i}^{\text{vapor}}=100\,{}^{\circ}\text{C}\). The mass of vapor, according to our assumptions, is the same mass of water and the same as ice, since no mass is lost during the processes, then \(m_{\text{vapor}}=m_{\text{water}}=m_{\text{ice}}\). Because all the masses are equal, we shall use the notation \(m\) for all masses. Putting everything together in equation \eqref{sup5}, we obtain

\begin{equation*}
Q_{\text{in}}=mc_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})+mL_{\text{f}}+mc_{\text{water}}(T_f^{\text{water}}-T_i^{\text{water}})\end{equation*} \begin{equation}\label{sup5.1}+mL_{\text{v}}+mc_{\text{vapor}}(T_{f}^{\text{vapor}}-T_i^{\text{vapor}}),
\end{equation}

where we have also changed \(m_{\text{water boils}}\) for \(m\) because all the water boils and \(m_{\text{ice melts}}=m\) because all the ice melts. We must now solve for \(T_f^{\text{vapor}}\) in equation \eqref{sup5.1}; thus,

\begin{equation*}
Q_{\text{in}}-mc_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})-mL_{\text{f}}-mc_{\text{water}}(T_f^{\text{water}}-T_i^{\text{water}})-mL_{\text{v}}\end{equation*} \begin{equation}=mc_{\text{vapor}}(T_{f}^{\text{vapor}}-T_i^{\text{vapor}}),
\end{equation}

dividing both sides by \(mc_{\text{vapor}}\), we obtain

\begin{equation*}
\frac{Q_{\text{in}}}{mc_{\text{vapor}}}-\frac{mc_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})}{mc_{\text{vapor}}}-\frac{mL_{\text{f}}}{mc_{\text{vapor}}}-\frac{mc_{\text{water}}(T_f^{\text{water}}-T_i^{\text{water}})}{mc_{\text{vapor}}}\end{equation*} \begin{equation}\label{sup5.2}-\frac{mL_{\text{v}}}{mc_{\text{vapor}}}=T_{f}^{\text{vapor}}-T_i^{\text{vapor}}.
\end{equation}

Simplifying the mass \(m\) in some of the terms and solving for \(T_{f}^{\text{vapor}}\) from equation \eqref{sup5.2}, we obtain

\begin{equation*}
T_{f}^{\text{vapor}}=T_i^{\text{vapor}}+\frac{Q_{\text{in}}}{mc_{\text{vapor}}}-\frac{c_{\text{ice}}(T_f^{\text{ice}}-T_i^{\text{ice}})}{c_{\text{vapor}}}-\frac{L_{\text{f}}}{c_{\text{vapor}}}\end{equation*} \begin{equation}-\frac{c_{\text{water}}(T_f^{\text{water}}-T_i^{\text{water}})}{c_{\text{vapor}}}-\frac{L_{\text{v}}}{c_{\text{vapor}}}.
\end{equation}

Using the numerical values for each variable, we finally get

\begin{equation*}
T_{f}^{\text{vapor}}=100\,{}^{\circ}\text{C}+\frac{1.2\times 10^7\,\text{J}}{(3\,\text{kg})(1996\,\text{J/kg}\,{}^{\circ}\text{C})}-\frac{2108\,\text{J/kg}\,{}^{\circ}\text{C}(-(-20\,{}^{\circ}\text{C}))}{1996\,\text{J/kg}\,{}^{\circ}\text{C}}\end{equation*} \begin{equation}-\frac{334000\,\text{J/kg}}{1996\,\text{J/kg}\,{}^{\circ}\text{C}}-\frac{4186\,\text{J/kg}\,{}^{\circ}\text{C}(100\,{}^{\circ}\text{C})}{1996\,\text{J/kg}\,{}^{\circ}\text{C}}-\frac{2.27\times 10^6\,\text{J/kg}}{1996\,\text{J/kg}\,{}^{\circ}\text{C}},
\end{equation}

namely,

\begin{equation}
T_f^{\text{vapor}}\approx 568\,{}^{\circ}\text{C}.
\end{equation}

Which is consistent with our assumption five \(T_f^{\text{vapor}}>100\,{}^{\circ}\text{C}\).

To explicitly answer the question asked in the problem, the final phase is vapor and its final temperature is \(568\,{}^{\circ}\text{C}\).

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