Ice Cubes in a Glass!

Using calorimetry, combine the heat equation for the three objects, and solve for the final temperature.

The amount of heat required for an object to increase its temperature is given by:

\begin{equation*}
Q = m c (T_f – T_i),
\end{equation*}

where in this case the heat exchange for the three objects involved (the water, the glass and the ice) can be written as:

\begin{equation*}
Q_w + Q_g + Q_{\text{ice}} = 0.
\end{equation*}

All the objects will have the final temperature as \(T_f\). Writing each heat as the first equation and solving for \(T_f\) after a lot of algebra we get:

\begin{equation*}
T_f = \frac{m_w c_w T_{i,w} + m_g c_g T_{i,g} + m_{\text{ice}} c_{\text{ice}} T_{i, ice}}{m_w c_w + m_g c_g + m_{\text{ice}} c_{\text{ice}}}.
\end{equation*}

Plugging in numerical values, we have:

\begin{equation*}
T_f = 6.53\, ^{\circ} \text{C}.
\end{equation*}

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

The problem asks us to find the final temperature of the system after it has reached thermal equilibrium. When the glass and the water come in contact with the ice cubes, there will be a net heat transfer between these objects until all of them reach the same temperature. In order to find the final temperature of the system, we need to relate the heat released or absorbed by all the components of the system. Then, we can relate heat with the temperature change of each material by using their masses and specific heat capacities. Finally, we can solve for the final temperatures from this expression.

First, we will assume that the system composed of the water, the glass, and the ice cubes is isolated. In other words, there is no heat exchange between these objects and the surroundings. This can be written as

\begin{equation}
\label{EQ:1}
Q = 0,
\end{equation}

where Q is the heat exchanged by all system components, and thus it can be written as

\begin{equation}
Q = Q_w + Q_g + Q_{\text{ice}},
\end{equation}

where \(Q_w, Q_g\) and \(Q_{\text{ice}}\) represent the heat exchanged by the water, the glass, and the ice, respectively. After inserting this in eq. \eqref{EQ:1}, we get

\begin{equation}
\label{EQ:2}
Q_w + Q_g + Q_{\text{ice}} = 0.
\end{equation}

This equation can be interpreted in the following way: since there is no heat exchange with the surrounding, there can only be heat exchange among the system components. In other words, if an object in the system releases heat, it must be absorbed by other objects in the system. Here, the heat released by the glass and the water will be absorbed by the ice cubes. Also, since both ice cubes are at the same temperature, we can consider them as a single component with its mass corresponding to the added mass of both ice cubes.

Now, we should relate the heat exchanged by each component with its temperature change. The amount of heat required for an object of mass \(m\) to absorb (or release) so that its temperature increases (or decreases) from \(T_i\) to \(T_f\) is given by

\begin{equation}
\label{EQ:3}
Q = m c (T_f – T_i),
\end{equation}

where \(c\) is the specific heat capacity of the object. It is a constant which depends on the type of material.

Now, we should write the heat exchanged by each item in terms of their initial and final temperatures using eq. \eqref{EQ:3}. Before jumping into that, notice that there will be heat exchange between the system components until they reach thermal equilibrium. At this point, their final temperatures will be the same. Hence, we will use \(T_f\) to denote the final temperature of all the components. This is the variable we should solve for.

Now, for the glass, eq. \eqref{EQ:3} can be written as

\begin{equation}
\label{EQ:4}
Q_g = m_g c_g (T_f – T_{i,g}),
\end{equation}

where \(m_g\) is the mass of the glass, \(c_g\) is its specific heat, and \(T_f\) and \(T_{i,g}\) are the glass temperatures after and before the water is added, respectively. Similarly, for the ice cubes we have

\begin{equation}
\label{EQ:5}
Q_{\text{ice}} = m_{\text{ice}} c_{\text{ice}} (T_f – T_{i, ice}),
\end{equation}

where \(m_{\text{ice}}\) represents the combined mass of both ice cubes, \(c_{\text{ice}}\) is the specific heat of ice (which is different than the specific heat of liquid water!), \(T_f\) is the final temperature of the system, and \(T_{i,ice}\) is the initial temperature of the ice cubes. Finally, for the water, we can write eq. \eqref{EQ:3} as

\begin{equation}
\label{EQ:6}
Q_w = m_w c_w (T_f – T_{i,w}),
\end{equation}

where \(m_w\) is the mass of the water, \(c_w\) is its specific heat, and \(T_{i,w}\) is its initial temperature, respectively.

If we substitute eqs. \eqref{EQ:4}, \eqref{EQ:4} and \eqref{EQ:6} in eq. \eqref{EQ:2}, we obtain

\begin{equation}
m_w c_w (T_f – T_{i,w}) + m_g c_g (T_f – T_{i,g}) + m_{\text{ice}} c_{\text{ice}} (T_f – T_{i, ice}) = 0.
\end{equation}

We should solve this equation for \(T_f\). If we distribute the products over the subtractions of temperatures, we obtain

\begin{equation}
m_w c_w T_f – m_w c_w T_{i,w} + m_g c_g T_f – m_g c_g T_{i,g} + m_{\text{ice}} c_{\text{ice}} T_f – m_{\text{ice}} c_{\text{ice}} T_{i, ice} = 0.
\end{equation}

Now, rearranging the equation so that only the terms containing \(T_f\) are on one side of the equation yields

\begin{equation}
m_w c_w T_f + m_g c_g T_f + m_{\text{ice}} c_{\text{ice}} T_f = m_w c_w T_{i,w} + m_g c_g T_{i,g} + m_{\text{ice}} c_{\text{ice}} T_{i, ice}.
\end{equation}

Factoring out \(T_f\) gives

\begin{equation}
(m_w c_w + m_g c_g + m_{\text{ice}} c_{\text{ice}}) T_f = m_w c_w T_{i,w} + m_g c_g T_{i,g} + m_{\text{ice}} c_{\text{ice}} T_{i, ice},
\end{equation}

and after dividing by the term in parentheses, we get

\begin{equation}
T_f = \frac{m_w c_w T_{i,w} + m_g c_g T_{i,g} + m_{\text{ice}} c_{\text{ice}} T_{i, ice}}{m_w c_w + m_g c_g + m_{\text{ice}} c_{\text{ice}}}.
\end{equation}

Now, notice that before the water was added, the ice cubes and the glass were in contact. Since no further information is given, we’ll assume that they were in contact for long enough that they were in thermal equilibrium when the water was added. Under this assumption, their initial temperatures are the same, i.e.

\begin{equation}
T_{i, ice} = T_{i,g} = -5\, ^{\circ} C.
\end{equation}

Additionally, the density of water is \(1 \frac{\text{g}}{\text{mL}}\). Therefore, the mass of 100 mL of water is \(100 \ \text{g}= 0.1 \ \text{kg},\). With this in mind, if we insert numerical values, we obtain

\begin{equation}
T_f = \frac{((0.1 \ \text{kg})(4180 \ \frac{\ \text{J}}{\ \text{kg} \, ^{\circ}{\text{C}}})(10 \, ^{\circ}{\text{C}}) + (0.1 \ \text{kg})(840\ \frac{\, \text{J} }{\ \text{kg} \, ^{\circ}{\text{C}}} \, ^{\circ}{\text{C}}))(-5 \, ^{\circ}{\text{C}}) + 2(0.01 \ \text{kg})(2093 \frac{\ \text{J}}{\ \text{kg} \, ^{\circ}{\text{C}}}))(-5 \, ^{\circ}{\text{C}}) )}{((0.1 \ \text{kg})(4180\ \ \frac{\ \text{J}}{\ \text{kg} \, ^{\circ}{\text{C}}} \, ^{\circ}{\text{C}})) + (0.1 \ \text{kg})(840\ \frac{\ \text{J}}{\ \text{kg} \, ^{\circ}{\text{C}}} \, ^{\circ}{\text{C}})) + 2(0.01 \ \text{kg})(2093\ \frac{\ \text{J}}{\ \text{kg} \, ^{\circ}{\text{C}}} \, ^{\circ}{\text{C}})},
\end{equation}

which is equivalent to

\begin{equation}
T_f = 6.53\, ^{\circ} \text{C}.
\end{equation}

As expected, the final temperature is between the initial temperature of the glass and ice cubes (-5 \(\, ^{\circ}\)C), and the initial temperature of water (10 \(\, ^{\circ}\)C).

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