# Heat of Fusion Problems and Solutions

The amount of energy needed to change a sample of a substance of mass $m$ from a solid to a liquid or vice versa without any change in its temperature is proportional to its mass with the following formula $Q_f=\pm mL_f$ where $L_f$ is called the latent heat of fusion of that substance.

Minus sign shows that heat must be removed from the substance during the freezing process.

The SI units of latent heat of fusion are $J/kg$ or $cal/kg$. It is also known as the enthalpy of fusion.

The latent heat of fusion of water is $333.5\,{\rm kJ/kg}$. This number indicates that to melt one kilogram of solid water (ice) into liquid water, we must extract $333.5\,{\rm kJ}$ heat from the ice.

In all heat problems, first, identify that whether there is a change in temperature or not. If there was then use the equation $Q=mc\Delta T$ to solve the unknown. If not, then read this article.

## Latent Heat of Fusion Example Problems with Answers:

Problem (1): How much heat is needed to change $2\,{\rm kg}$ of ice at ${\rm -20^\circ C}$ into ${\rm 100^\circ C}$ water? (specific heats of ice and water are $2.05\,{\rm kJ/kg\cdot ^\circ \!C}$ and $4.18\,{\rm kJ/kg\cdot ^\circ \!C}$, respectively)

Solution: this is the simplest example of a latent heat problem. The total heat needed to change ice into steam breaks into four parts: $Q_1$, is the heat required to change its temperature from ${\rm -20^\circ C}$ to ${\rm 0^\circ C}$. $Q_2$ is the heat needed to melt the ice or phase change. $Q_3$ is the heat required to increase the temperature from $0^\circ C$ to $100^\circ C$ and $Q_4$ is the heat needed to vaporize the water. Now, we calculate each part as follows:

In the part one, there is a change in temperature so we have \begin{align*}Q_1&=mc_{ice}(T_f-T_i)\\&=(2)(2.05)(0-(-20))\\&=82\quad {\rm kJ}\end{align*}In the next level, there is a phase change (solid to liquid) so we have \begin{align*} Q_2&=mL_f\\&=(2)(333.5)\\&=667\quad {\rm kJ}\end{align*}In the final level, the temperature changes so \begin{align*}Q_3&=mc_{water}\Delta T\\&=(2)(4.18)(100-0)\\&=836\quad{\rm kJ}\end{align*}By adding the above heats together, the total heat required is found $Q=Q_1+Q_2+Q_3=1585\,{\rm kJ}$

Problem (2): How much heat is needed to transform $2\,{\rm kg}$ of ice at ${\rm -15 ^\circ C}$ into $2\,{\rm kg}$ of water at ${\rm 70^\circ C}$?

Solution: first, the ${\rm -15^\circ C}$-ice must absorb the following heat $Q_1$ to reaches ${\rm 0^\circ C}$-ice. \begin{align*}Q_1&=mc_{ice}\Delta T\\&=(2)(2.05)(0-(-15))\\&=61.5\quad{\rm kJ}\end{align*}Then, it must transform into ${\rm 0^\circ C}$-water by absorbing the heat $Q_2$ as below \begin{align*} Q_2&=mL_f\\&=(2)(333.5)\\&=667\quad{\rm kJ}\end{align*}In the final stage, the heat $Q_3$ is needed to increase the temperature of the water (melted ice) from ${\rm 0^\circ C}$ to ${\rm 70^\circ C}$ as following \begin{align*} Q_3&=mc_{water}\Delta T\\&=(2)(4.18)(70-0)\\&=585.2\quad {\rm kJ}\end{align*}Summing up the above heats get the total heat needed to ${\rm -15^\circ C}$-ice absorbs and transforms into ${\rm 70^\circ C}$-water. $Q=Q_1+Q_2+Q_3=1313.7\quad {\rm kJ}$ As you can see, the heat required to melt the ice is the smallest amount. Since in the solid phase (ice) the molecules are very close together thus given heat quickly spread across the whole solid and consequently much less heat is needed.

Problem (3): In an isolated container, we add $4\,{\rm kg}$ of ice at ${\rm -10^\circ C}$ to the $40$ liters of existing water at ${\rm 40^\circ C}$. When the water reaches equilibrium, what is its temperature?

Solution: here, we have two materials: ice gains heat, warms, and then melts. Water releases its heat (because of its higher temperature) and cools down to a lower temperature. To see, whether the heat released by water can melt the whole ice we must first find the gained and released heat as below:

Ice gains the heat $Q_1+Q_2$ to reaches ${\rm 0^\circ C}$-water where $Q_1$ is for ${\rm -10^\circ C}\rightarrow {\rm 0^\circ C}$ and $Q_2$ is for ${\rm 0^\circ C}$-ice to ${\rm 0^\circ C}$-water, thus \begin{align*}Q_1&=m c_{ice}\Delta T\\&=(4)(2.05)(0-(-10)\\&=82\quad {\rm kJ}\\ \\ Q_2&=mL_f\\ &=(4)(333.5)\\&=1334\quad{\rm kJ}\end{align*}Consequently, the energy gained by the ice is $Q_{gained}=Q_1+Q_2=1416\,{\rm kJ}$ Next, we must find how much heat energy $Q_3$ is released during the water cooling process: ${\rm 40^\circ C}$-water to ${\rm 0^\circ C}$-water. Keep in mind that in the problem the volume of the water is given so using the definition of density, $\rho=mV$, we can find its mass as below \begin{align*} m&=\rho V\\ \\ &={\rm \left(1000\,\frac{kg}{m^3}\right)(40\,L)\left(\frac{1\,m^3}{1000\,L}\right)}\\ \\&=40\quad {\rm kg}\end{align*}Where the last parenthesis is the conversion factor from $L\rightarrow m^3$.

Now we compute the heat $Q_3$ as below \begin{align*}Q_3&=mc_{water}\Delta T\\&=(\rho V)c\Delta T\\&=(40)(4.18)(0-40))\\&=-6688\quad {\rm kJ}\end{align*} The minus sign indicates that the heat is leaving the water.

The comparison between those two heats shows that the heat released by the water is enough to melt the ice totally (reaches the ${\rm 0^\circ C}$-water) and its remaining causes the temperature of the water to reach the final point $T_f$.

Thus, the remaining heat $Q_{net}=6688-1416=5272\,{\rm kJ}$ changes the temperature of $40\,{\rm kg}$ of water at ${\rm 0^\circ C}$ to some temperature $T_f$ as below \begin{align*} Q_{net}&=mc_{w}\Delta T\\5272&=(40)(4.18)(T_f-0)\\\Rightarrow T_f&={\rm 31.53^\circ C}\end{align*}
In the above, we used the absolute value of the heat released by the water. Th negative is a sign of the input or the output of heat and has no effect on solving the heat problems.

There is also another solution for this example problem with the help of conservation of energy. See the next problem.

Problem (4): A $200-g$ ice cube at ${\rm 0^\circ C}$ is dropped into ${\rm 1-kg}$ of water initially at ${\rm 10^\circ C}$. What is the final temperature of the mixture? (Assume no heat is released to the surroundings)

Solution
Method (I): the heat $Q_{gain}$ is the absorbed heat by the ice to transform from ${\rm 0^\circ C}$-ice to ${\rm 0^\circ C}$-water as below \begin{align*}Q_{gained}&=m_{ice}L_f\\ &={\rm (0.200\,kg)(333.5\,kJ/kg)}\\ &=66.7\quad{\rm kJ}\end{align*} The heat released by the $1\,{\rm kg}$ of water to transform from ${\rm 10^\circ C}\rightarrow {\rm 0^\circ C}$ is calculated as below \begin{align*}Q_{lost}&=m_{w}c\Delta T\\&=(1)(4.18)(0-10)\\&=-41.8\quad {\rm kJ}\end{align*} As you can see, ice needs more heat to completely melts and transforms into ${\rm 0^\circ C}$-ice since $Q_{gained}>|Q_{lost}|$ Where $|\cdots|$ gets the absolute value of the heat lost. Thus, the ice can not be totally melted and there be some ice cube in the mixture.

Method (II): using conservation of energy

At first, we assume that all of the ice melts and transforms at ${\rm 0^\circ C}$ into the water at temperature $T_f$. Ice gains the thermal energy released by the water.

Since there is no heat exchange with the environment so according to the conservation of energy sum of those heats must be zero. Therefore, \begin{gather*} Q_{gained}+Q_{lost}=0 \\ \\ \underbrace{m_{ice}L_f}_{Q_1}+\underbrace{m_{ice}c_{ice}(T_f-0)}_{Q_2}+\underbrace{m_wc_w(T_f-10)}_{Q_3}=0\\ \\ (0.200)(333.5)+(0.200)(2.09)(T_f-0)+(1)(4.18)(T_f-10)=0 \\ \\ \Rightarrow T_f={\rm -5.5^\circ C}\end{gather*} A final temperature of below ${\rm 0^\circ C}$ is not correct because no amount of ice at ${\rm 0^\circ C}$ can decrease the temperature of a substance at higher temperature $T$ to below ${\rm 0^\circ C}$.

As a rule of thumb, in all heat problems, the final (equilibrium) temperature is always between the initial temperatures of the given materials.

Consequently, at the final, we have a mixture of ice and water with the equilibrium temperature of ${\rm 0 ^\circ C}$.

Problem (5): A ${\rm 80-g}$ ice cube at ${\rm 0^\circ C}$ is dropped into ${\rm 725-g}$ of water originally at ${\rm 24^\circ C}$. What is the final temperature of the mixture? ($c_{water}=4.18\,{\rm kJ/kg\cdot ^\circ C}$ , $c_{ice}=2.090\,{\rm kJ/kg\cdot ^\circ C}$, and $L_f=333.5\,{\rm kJ/kg}$).

Solution: water has higher temperature so it releases its heat as $Q_{released}=Q_1$ and the ice absorbs it $Q_{abs}=Q_2+Q_3$. Initially, we assume that the all of the ice melts and converts to ${\rm 80-g}$ of water at temperature of $T_f$. By conservation of energy, the algebraic sum of those heats must be zero. Thus, we have \begin{gather*} Q_{released}+Q_{gained}=0\\ \\ \underbrace{m_w c_w(T_f-24)}_{Q_1}+\underbrace{m_{ice}L_f}_{Q_2}+\underbrace{m_{ice}c_{ice}(T_f-0)}_{Q_3}=0\\ \\ (0.725)(4.180)(T_f-24)+(0.080)(333.5)+(0.080)(4.18)(T_f-0)=0\\ \\ -(3.0305)(T_f-24)+26.68=0\\ \\ \Rightarrow T_f= {\rm 13.6^\circ C}\end{gather*} Therefore, at the final the equilibrium temperature is ${\rm 13.6^\circ C}$.

Problem (6): Into a aluminum container with mass of ${\rm 400-g}$ and specific heat of $c_{Al}=0.900\,{\rm kJ/kg\cdot ^\circ C}$ at ${\rm 25^\circ C}$ is poured ${\rm 5\,kg}$ of ethyl alcohol at ${\rm 30^\circ C}$ and is dropped a ${\rm 200\,g}$ ice cube at ${\rm -15^\circ C}$. Assume the system is isolated from the surrounding. What is the final equilibrium temperature of the mixture?

Solution: Since the container is isolated, so there is no heat exchange with the surrounding. By conservation of energy, some of the heats gained and lost must be zero.

Aluminum container and alcohol lose their thermal energies since they are at a higher temperature and the ice gains those heat.

As before, we assume that all of the ice melts and reaches at temperature of $T_f$ higher than zero. During this process $\underbrace{-15^\circ C}_{ice}\stackrel{Q_1}{\longrightarrow} \underbrace{0^\circ C}_{ice}\stackrel{Q_2}{\longrightarrow}\underbrace{0^\circ C}_{water}\stackrel{Q_3}{\longrightarrow} \underbrace{T_f}_{water}$ the amount of thermal heats gained by the ice is computed as below \begin{align*} Q_1&=m_{ice}c_{ice}(0-(-15))\\&=(0.200)(2.090)(15)\\&=6.27\quad {\rm kJ} \\ \\ Q_2&=m_{ice}L_f\\&=(0.200)(333.5)\\&=66.7\quad {\rm kJ} \\ \\ Q_3&=m_{ice}c_{water}(T_f-0)\\&=(0.200)(4.18)(T_f)\end{align*} Therefore, the total thermal energy gained is \begin{align*} Q_{gained}&=Q_1+Q_2+Q_3\\&=72.97+0.836T_f\end{align*} On the other side, the container and the water release their thermal heats and reaches at final temperature of $T_f$ below their initial temperatures as  \begin{align*}Q_{lost}&=Q_{Al}+Q_{ethyl}\\ \\&=m_{Al}c_{Al}(T_f-25)+m_{ethyl}c_{ethyl}(T_f-30)\\ \\&=(0.400)(0.9)(T_f-25)+(5)(2.4)(T_f-30)\\\\&=12.36T_f-369\end{align*} According to the principle of the conservation of energy all heat lost must be absorbed by the ice so \begin{align*} Q_{lost}+Q_{gained}&=0\\\\ \Rightarrow\quad Q_{gained}&=-Q_{lost} \\ \\ 72.97+0.836T_f&=-(12.36T_f-369)\\ \\ \Rightarrow \quad T_f&={\rm 22.43^\circ C}\end{align*}

Problem (7): How much thermal energy does a freezer have to remove from ${\rm 2\, kg}$ of water at a temperature of ${\rm 25^\circ C}$ to make ice at ${\rm -15^\circ C}$? (Assume the ice specific heat capacity is $c_{ice}=2.090\,{\rm kJ/kg\cdot ^\circ C}$ and the water is $c_{w}=4.180\,{\rm kJ/kg\cdot ^\circ C}$).

Solution: the process involved is outlined in the below $\underbrace{25^\circ C}_{water}\stackrel{Q_1}{\longrightarrow}\underbrace{0^\circ C}_{water}\stackrel{Q_2}{\longrightarrow}\underbrace{0^\circ C}_{ice}\stackrel{Q_3}{\longrightarrow}\underbrace{-15^\circ C}_{ice}$ Using the definition of heat capacity, $Q=mc\Delta T$, and heat of fusion, $Q=mL_f$, we have, \begin{align*}Q_1&=mc_w(0-25)\\ &=(2)(4.18)(-25)\\&=-209\quad {\rm kJ}\\ \\ Q_2&=-mL_f\\&=-(2)(333.5)\\&=-667\quad {\rm kJ}\\ \\ Q_3&=mc_{ice}(-15-0)\\&=(2)(2.090)(-15)=-62.7\quad {\rm kJ}\end{align*} Next, summing those heats get the total thermal energy must be removed from the water to lower its temperature from ${\rm 25^\circ C}$ to ${\rm -15^\circ C}$. \begin{align*} Q_{tot}&=Q_1+Q_2+Q_3\\ &=-209-667-62.7\\&=-938.7\quad {\rm kJ}\end{align*} Note that the stage two is a freezing process so we must use the definition of the heat of fusion as $Q=-mL_f$ since the heat must be removed from the water.

In all heat problems, the negatives show that the thermal energy flows out of the system as we expected.

Problem (8): A chunk of ice is taken from the freezer at ${\rm -15^\circ C}$ and is dropped into  ${\rm 200\,g}$-an aluminum container filled with ${\rm 450\,g}$ of water at a temperature of ${\rm 24^\circ C}$. After a while, all we have left is water at ${\rm 5^\circ C}$. What was the mass of the ice cubes?

Solution: first identify which material loses its heat and which gains it. Because the container and the water inside it are at a higher temperature than the ice cubes, so they release their heat and ice absorbs it.

The process of reaching ${\rm -15^\circ C}$-ice to water at ${\rm 5^\circ C}$ is as below $\underbrace{{\rm -15^\circ C}}_{ice}\stackrel{Q_1}{\longrightarrow}\underbrace{{\rm 0^\circ C}}_{ice}\stackrel{Q_2}{\longrightarrow}\underbrace{{\rm 0^\circ C}}_{water}\stackrel{Q_3}{\longrightarrow}\underbrace{{\rm 5^\circ C}}_{water}$ By applying definitions of specific heat and latent heat of fusion the heats are determined as below \begin{align*} Q_1&=m_{ice}c_{ice}(0-(-15))\\&=m_{ice}(2.090)(15)\\&=31.35m_{ice}\\ \\ Q_2&=m_{ice}L_f\\&=m_{ice}(333.5)\\ \\Q_3&=m_{ice}c_{water}(5-0)\\&=m_{ice}(4.180)(5)\\&=20.9m_{ice}\end{align*} Summing those get the absorbed (gained) heat by the ice $Q_{abs}=385.75m_{ice}$ Next, the container and the water inside it release the heat $Q_{lost}$ and reach from ${\rm 24^\circ C}$ to ${\rm 5^\circ C}$ as below \begin{align*} Q_{lost}&=(m_{cont}c_{Al}+m_{water}c_{water})(5-24)\\&=(0.2\times 0.9+0.450\times 4.18)(-19)\\ &=-39.159\quad {\rm kJ}\end{align*} A negative heat obtained as we expected for a heat lost in a system. Now, setting the heat lost equal to the heat absorbed (conservation of energy) and solving for the unknown mas $m_{ice}$, we get \begin{align*} \text{gained heat}&=-\text{lost heat} \\ 385.75m_{ice}&=-(-39.159) \\ \Rightarrow \quad m_{ice}&=0.101\quad {\rm kg}\end{align*} Therefore, a ${\rm 101\,g}$-chunk of ice can lower the temperature of the water and the container to the ${\rm 5^\circ C}$.

For solving more problems you can also check out the following page:
Thermodynamics problems with solutions for AP physics

Date Published: 4/11/2021
Author: Ali Nemati