In a calorimeter can (which behaves thermally as if it were...

In a calorimeter can (which behaves thermally as if it were equivalent to $40 \mathrm{~g}$ of water) are $200 \mathrm{~g}$ of water and $50 \mathrm{~g}$ of ice, all at exactly $0{ }^{\circ} \mathrm{C}$. Into this is poured $30 \mathrm{~g}$ of water at $90{ }^{\circ} \mathrm{C}$. What will be the final condition of the system? Let us start by assuming (perhaps incorrectly) that the final temperature is $T_{f}>0{ }^{\circ} \mathrm{C}$. Then $$ \begin{array}{c} \left(\begin{array}{l} \text { Heat change of } \\ \text { hot water } \end{array}\right)+\left(\begin{array}{l} \text { Heat to } \\ \text { melt ice } \end{array}\right)+\left(\begin{array}{l} \text { Heat to warm } \\ 250 \mathrm{~g} \text { of water } \end{array}\right)+\left(\begin{array}{l} \text { Heat to warm } \\ \text { calorimeter } \end{array}\right)=0 \\ (30 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-90{ }^{\circ} \mathrm{C}\right)+(50 \mathrm{~g})(80 \mathrm{cal} / \mathrm{g})+(250 \mathrm{~g})\left(1 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-0{ }^{\circ} \mathrm{C}\right) \\ +(40 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-0{ }^{\circ} \mathrm{C}\right)=0 \end{array} $$ Solving gives $T_{f}=-4.1^{\circ} \mathrm{C}$, contrary to our assumption that the final temperature is above $0{ }^{\circ} \mathrm{C}$. Apparently, not all the ice melts. Therefore, $T_{f}=0{ }^{\circ} \mathrm{C}$. To find how much ice melts, we write Heat lost by hot water $=$ Heat gained by melting ice $$ (30 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(90{ }^{\circ} \mathrm{C}\right)=(80 \mathrm{cal} / \mathrm{g}) m $$ where $m$ is the mass of ice that melts. Solving this equation yields $m=34 \mathrm{~g}$. The final system has $50 \mathrm{~g}-34 \mathrm{~g}=16 \mathrm{~g}$ of ice not melted.

In a calorimeter can (which behaves thermally as if it were equivalent to $40 \mathrm{~g}$ of water) are $200 \mathrm{~g}$ of water and $50 \mathrm{~g}$ of ice, all at exactly $0{ }^{\circ} \mathrm{C}$. Into this is poured $30 \mathrm{~g}$ of water at $90{ }^{\circ} \mathrm{C}$. What will be the final condition of the system?
Let us start by assuming (perhaps incorrectly) that the final temperature is $T_{f}>0{ }^{\circ} \mathrm{C}$. Then
$$
\begin{array}{c}
\left(\begin{array}{l}
\text { Heat change of } \\
\text { hot water }
\end{array}\right)+\left(\begin{array}{l}
\text { Heat to } \\
\text { melt ice }
\end{array}\right)+\left(\begin{array}{l}
\text { Heat to warm } \\
250 \mathrm{~g} \text { of water }
\end{array}\right)+\left(\begin{array}{l}
\text { Heat to warm } \\
\text { calorimeter }
\end{array}\right)=0 \\
(30 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-90{ }^{\circ} \mathrm{C}\right)+(50 \mathrm{~g})(80 \mathrm{cal} / \mathrm{g})+(250 \mathrm{~g})\left(1 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-0{ }^{\circ} \mathrm{C}\right) \\
+(40 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(T_{f}-0{ }^{\circ} \mathrm{C}\right)=0
\end{array}
$$
Solving gives $T_{f}=-4.1^{\circ} \mathrm{C}$, contrary to our assumption that the final temperature is above $0{ }^{\circ} \mathrm{C}$. Apparently, not all the ice melts. Therefore, $T_{f}=0{ }^{\circ} \mathrm{C}$.
To find how much ice melts, we write
Heat lost by hot water $=$ Heat gained by melting ice
$$
(30 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(90{ }^{\circ} \mathrm{C}\right)=(80 \mathrm{cal} / \mathrm{g}) m
$$
where $m$ is the mass of ice that melts. Solving this equation yields $m=34 \mathrm{~g}$. The final system has $50 \mathrm{~g}-34 \mathrm{~g}=16 \mathrm{~g}$ of ice not melted.

Key Concepts

Conservation of Energy in Thermal Processes

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In thermal processes, this means that the total heat lost by warmer components must equal the heat gained by cooler components or phase change processes. This concept provides the fundamental framework for setting up and solving energy balance equations in calorimetry problems.

Calorimetry

Calorimetry is the study of measuring the change in heat during physical and chemical processes. It involves using a calorimeter, which is a device designed to minimize heat exchange with the environment, thereby allowing accurate determination of energy changes in a system. Calorimetry is fundamental in fields like thermodynamics and chemistry as it helps quantify the heat transferred during processes such as chemical reactions, phase changes, and thermal mixing.

Specific Heat Capacity

Specific heat capacity is a material property that indicates the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. This concept is critical in calculating the temperature changes in systems when heat is absorbed or released, as it relates directly to the mass of the substance and the magnitude of the temperature change.

Latent Heat of Fusion

Latent heat of fusion is the amount of heat energy needed to change a substance from solid to liquid at its melting point without changing its temperature. This concept is crucial when analyzing processes involving phase changes, as the energy associated with the breaking of intermolecular bonds during melting must be accounted for in energy balances.

Thermal Equilibrium

Thermal equilibrium refers to the state where all parts of a system achieve the same temperature, resulting in no net transfer of heat energy among them. In problems involving mixtures of substances at different temperatures, determining the final equilibrium temperature is a key objective and requires careful application of energy conservation principles.

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