# Heat of Vaporization Problems with Answers

When a substance changes its state from a liquid to steam or vice versa, the heat absorbed or released during this process does not lead to a change in the temperature of the substance.

In physics the amount of heat lost or gained by a substance when it changes from a gas to liquid (condensing) or liquid to gas (boiling) depends on the total mass $m$ of the substance and its latent heat of vaporization $L_v$ whose formula is $Q=\pm mL_v$ where plus ($+$) is for a change from liquid to gas and negative ($-$) is for gas to liquid.

The SI unit of latent heat of vaporization is joule per kilogram $J/kg$.

For example, The latent heat of vaporization of water is 2260 kJ/kg that indicates the heat required to transform 1 kg of water completely into 1 kg of steam is 2260 kJ.

In the following example problems, by applying latent heat of vaporization, a number of heat problems involving phase change are answered.

## Heat of Vaporization Problems:

Problem (1): A 10-g chunk of liquid lead at 1750°C takes 8580 joules of heat to turns into 10-g of gaseous lead at 1750°C. What is the latent heat of vaporization of the lead?
(a) 858 kJ/kg        (b) 296 kJ/kg        (c) 4726 kJ/kg        (d) 879 kJ/kg

Solution: a liquid stage is being converted into a gaseous state so there is phase change (boiling). Substitute the known values into the heat of vaporization formula (with plus sign) as below \begin{align*} Q&=+mL_v\\ rm 8580\, {\rm J}&=(0.010\,{\rm kg})L_v \\ \Rightarrow L_v&=858\quad {\rm kJ/kg}\end{align*} Thus, the heat of vaporization of the lead is 858 joules per kilogram or the correct answer is (a)

Problem (2): How much thermal energy needed to change a $30-g$ ice initially at ${\rm -5^\circ C}$ to steam at ${\rm 110^\circ C}$? (Assume $c_{ice}=2.090\,{\rm kJ/kg\cdot ^\circ C}$, $c_{water}=4.18\,{\rm kJ/kg\cdot ^\circ C}$, $c_{steam}=2.010\,{\rm kJ/kg\cdot ^\circ C}$, $L_f=333.5\,{\rm kJ/kg}$ and $L_v=2260\,{\rm kJ/kg}$)

(a) 91 kJ        (b) 102 kJ        (c) 84 kJ        (d) 125 kJ

Solution: The path from ice at ${\rm -5^\circ C}$ to steam at ${\rm 110^\circ C}$ is as following $\underbrace{\rm -5^\circ C}_{ice}\stackrel{Q_1}{\longrightarrow}\underbrace{\rm 0^\circ C}_{ice}\stackrel{Q_2}{\longrightarrow}\underbrace{\rm 0^\circ C}_{water}\stackrel{Q_3}{\longrightarrow}\underbrace{\rm 100^\circ C}_{water}\stackrel{Q_4}{\longrightarrow}\underbrace{\rm 100^\circ C}_{steam}\stackrel{Q_5}{\longrightarrow}\underbrace{\rm 110^\circ C}_{steam}$ The amount of heats are calculated as below
\begin{align*} Q_1&=mc_{ice}(0-(-5))\\&=(0.030)(2.090)(5)\\&=0.3135 \quad {\rm kJ}\\ \\Q_2&=mL_f\\&=(0.030)(333.5)\\&=10.005 \quad {\rm kJ}\\ \\Q_3&=mc_{water}(100-0)\\&=(0.030)(4.18)(100)\\&=12.54 \quad {\rm kJ}\\ \\Q_4&=mL_v\\&=(0.03)(2260)\\&=67.8 \quad {\rm kJ}\end{align*} For  the phase change occurred, the amount of heat required $Q_4$ was obtained using latent heat of vaporization formula $Q_v=mL_v$.

Finally, adding more thermal energy leads increasing in the temperature of the steam whose amount is determined as below \begin{align*} Q_5&=mc_{steam} \Delta T\\&=(0.030)(2.010)(110-100)\\&= 0.603\quad {\rm kJ}\end{align*}
In the above, there are two-phase changes, one is due to the ice $\stackrel{Q_2}{\rightarrow}$ water, and the other is water $\stackrel{Q_4}{\rightarrow}$ steam.

The total heat absorbed by the ice is the sum of the partial heats above $Q_{tot}=91.2\quad {\rm kJ}$ so the correct answer is (a).

To better understand the physics of the melting stage ($Q_2$), check out the following page:
Heat of fusion Problems with answers

Problem (3): How much heat must be released by the 200 g of steam at ${\rm 175^\circ C}$ to turns into 200 g of water at ${\rm 0^\circ C}$?

(a) 283 kJ        (b) 326 kJ         (c) 483 kJ         (d) 525 kJ

Solution: Here, steam wants to change its phase and turns into water (condense). The path for this process is as follows:

$\underbrace{\rm 175^\circ C}_{steam}\stackrel{Q_1}{\longrightarrow}\underbrace{\rm 100^\circ C}_{steam}\stackrel{Q_2}{\longrightarrow}\underbrace{\rm 100^\circ C}_{water}\stackrel{Q_3}{\longrightarrow}\underbrace{\rm 0^\circ C}_{water}$

In the first stage, there is a change in the temperature of the steam so we use the specific heat formula as below to find the heat $Q_1$ \begin{align*} Q_1&=m_s c_s \Delta T\\&=(0.200)(2010)(100-175)\\&=-30150\quad {\rm J}\\&=-30.15\quad {\rm kJ}\end{align*}

In the second stage, the steam cools and turns into the water so there is a phase change and the required thermal energy is found using latent heat of vaporization formula below \begin{align*}Q_2&=-mL_V\\&=-{\rm (0.200\, kg)\left(2260\,kJ/kg\right)}\\&=-452\quad {\rm kJ}\end{align*} Note that since the steam has lost its heat so we must add a negative in front of latent heat of vaporization as above.

In the third stage, the water cools and its temperature drops so we must use the specific heat capacity formula to find the heat lost during this stage, \begin{align*} Q_3&=m_s c_{w} \Delta T\\ &={\rm (0.200\,kg)(4.186\,kJ/kg\cdot ^\circ C)(0-100)}\\&=-0.8372\,{\rm kJ}\end{align*} Now, sum the above heats lost to find the total heat lost by the steam to convert from ${\rm 175^\circ C}$ to ${\rm 0^\circ C}$. $Q_{tot}=483\quad {\rm kJ}$ Thus, the correct answer is (c).

Problem (4): A ${\rm 10-g}$ chunk of ice at ${\rm -10^\circ C}$ is placed in a sealed container so that there is no heat exchange with the environment. How much heat is required to change it into steam at ${\rm 180^\circ C}$?

(a) 23.3 kJ         (b) 30.3 kJ         (c) 54.1 kJ         (d) 13.2 kJ

Solution: As above, the process is as follows $\underbrace{\rm -10^\circ C}_{ice}\stackrel{Q_1}{\longrightarrow}\underbrace{\rm 0^\circ C}_{ice}\stackrel{Q_2}{\longrightarrow}\underbrace{\rm 0^\circ C}_{water}\stackrel{Q_3}{\longrightarrow}\underbrace{\rm 100^\circ C}_{water}\stackrel{Q_4}{\longrightarrow}\underbrace{\rm 180^\circ C}_{steam}$ Use the equation $Q=mc\Delta T$ to find the heat needed to increase the temperature of the ice from ${\rm -10^\circ C}$ to ${\rm 0^\circ C}$ \begin{align*}Q_1&=mc_{ice}(0-(-10))\\&=(0.010)(2.090)(10)\\&=0.2090\quad {\rm kJ}\end{align*} There is a phase change in the next stage as the ice wants to turns into its liquid phase that is water. Thus, use the latent heat of fusion formula, $Q=mL_f$, to find the required heat \begin{align*} Q_2&=mL_f\\&=(0.010)(333.5)\\&=3.335\quad {\rm kJ}\end{align*} In the third stage, the additional heat would causes the temperature of the water increases. Thus, \begin{align*}Q_3&=mc_{water}(100-0)\\&=(0.010)(4.18)(100)\\&=4.18\quad {\rm kJ}\end{align*} In the final stage, the water wants to turn from liquid state into gas state so there is a phase change and we must use the latent heat of vaporization to find the needed heat as below \begin{align*}Q_4&=mL_v\\&=(0.010)(2260)\\&=22.6\quad {\rm kJ}\end{align*} Summing up the above heats, we get the total thermal energy required as $Q_{tot}=30.315\quad {\rm kJ}$ The correct answer is (b).

For the third stage, you can solve more problems on the following page:
Practice problems on the specific heat

Problem (5): 20-g of aluminum at ${\rm 200^\circ C}$ and 35-g of lead are dropped into a ${\rm 45\,cm^3}$ container full of ethyl alcohol at ${\rm 15^\circ C}$. After a while, the mixture reaches equilibrium with temperature of ${\rm 25^\circ C}$. What was the initial temperature of the lead? (density and specific heat capacity of ethyl alcohol are $\rho_{ethyl}={\rm 790\,kg/m^3}$, $c_{ethyl}={\rm 2400\,J/kg\cdot ^\circ C}$, respectively. $c_{Al}=900\,{\rm J/kg\cdot \circ C}$, $c_{Pb}=128\,{\rm J/kg\cdot \circ C}$).

(a) -230°C         (b) -487°C         (c) +487°C         (d) +230°C

Solution: First of all, using definition of density find the mass of the ethyl alcohol in the container. \begin{align*}m&=\rho V\\&={\rm \left(790\,\frac{kg}{m^3}\right)(45\times 10^{-6}\,m^3)}\\&=0.035550\quad{\rm kg}\\&=35.55\quad{\rm g}\end{align*}

In all latent heat problems, one substance lost its thermal energy and the other absorbs it (assuming there is no heat exchange with the environment). In such processes, the conservation of energy states that heat lost must be equal to the heat absorbed.

In this problem, the equilibrium temperature is higher than the temperature of the ethyl alcohol so it must gain heat whose value is obtained as below \begin{align*}Q_{gained}=m_{ethyl}c_{ethyl}(T_f-T_i)\\&=(0.03555)(2400)(25-15)\\&=853.2\quad {\rm J}\end{align*}

Aluminum must release the heat to decrease its temperature. Therefore, \begin{align*}Q_{lost-Al}&=m_{Al}c_{Al}(T_f-T_i)\\&=(0.020)(900)(25-200)\\&=-3150\quad {\rm J}\end{align*} The initial temperature of the lead is unknown, so we can assume it is higher than the final temperature of the mixture. Thus, the amount of heat lost by the lead is as below \begin{align*}Q_{lost-Pb}&=m_{Pb}c_{Pb}(T_f-T_i)\\&=(0.035)(128)(25-T_i)\\&=4.48(25-T_i)\end{align*} Applying conservation of energy as below and solving for the initial temperature of the lead, we have
\begin{align*}{-heat \ lost}&={heat \ gained}\\ \\ -(-3150+4.48(25-T_i))&=853.2\\ \\ 4.48(25-T_i)&=4003.2 \\ \\ \Rightarrow T_i&=-{\rm 487^\circ C}\end{align*} As you can see, the lead's initial temperature obtained much more below the equilibrium temperature of the mixture which contradicts the original assumption, $T_i>{\rm 25^\circ C}$.

Now, assume $T_i< {\rm 25^\circ C}$ so the lead must gain heat. Applying again the conservation of energy, we get \begin{align*}{-heat \ lost}&={heat\ gained}\\ \\ -(-3150)&=853.2+4.48(25-T_i)\\ \\ \Rightarrow T_i&=-{\rm 487^\circ C}\end{align*} which is consistent with the initial assumption. Therefore, the correct answer is (b).

Problem (6): In a 100-g glass container, there is 200 g of water initially at ${\rm 27^\circ C}$. How much steam at ${\rm 150^\circ C}$ is needed to raise the temperature of the container and water inside it to ${\rm 45^\circ C}$? ($c_{w}=4186\,{\rm J/kg\cdot ^\circ C}$, $c_{g}=837\,{\rm J/kg\cdot ^\circ C}$, heat of vaporization of water is $2260\,{\rm kJ/kg}$).

(a) 5.5 g         (b) 3.8 g         (c) 4.5 g         (d) 6.5 g

Solution: First of all, we calculate the amount of heat required to raise the temperature of container+water from ${\rm 27^\circ C}$ to ${\rm 45^\circ C}$ as below \begin{align*} Q_{gain}&=(m_{g}c_{g}+m_{w}c_{w})\Delta T\\ &=(0.100\times 837+0.200\times 4186)(45-27)\\&=16576.2\quad {\rm J}\end{align*} In above the subscript $g$ and $w$ indicate glass'' and water'', respectively. Thus, the combination of water+container must absorb $16.576\,{\rm kJ}$ heat to increase its temperature.

The amount of heat released by the steam to reach from ${\rm 150^\circ C}$ to ${\rm 45^\circ C}$ is composed of the following processes $\underbrace{{\rm 150^\circ C}}_{steam}\stackrel{Q_1}{\longrightarrow}\underbrace{{\rm 100^\circ C}}_{steam}\stackrel{Q_2}{\longrightarrow}\underbrace{{\rm 100^\circ C}}_{water}\stackrel{Q_3}{\longrightarrow}\underbrace{{\rm 45^\circ C}}_{water}$ In the third stage, the steam changes phase and turns into the water. The heat $Q_1$ is found as \begin{align*} Q_1&=m_s c_s \Delta T\\&=m_s (2.09)(100-150)\\&=-104.5m_s\quad {\rm kJ}\end{align*} The heat $Q_2$ is due to the phase change so using latent heat of vaporization formula, we have \begin{align*} Q_2&=-m_{s} L_v\\&=-m_s (2260)\quad {\rm kJ}\end{align*} The heat $Q_3$ causes a change in temperature so we have \begin{align*} Q_3&=m_s c_w \Delta T\\&=m_s (4.186)(45-100)\\&=-230.23m_s \quad {\rm kJ}\end{align*} Summing those heats get the thermal energy lost by the steam as $Q_{lost}=-2594.73m_s\quad{\rm kJ}$  According to conservation of energy heat lost must be equal with the heat gained \begin{align*} -{lost\ heat}&={gained\ heat}\\ 2594.73\,m_s&=16.576 \\ \Rightarrow m_s&=0.0063\quad {\rm kg}\end{align*} Thus, about $6.5\,{\rm g}$ of steam at ${\rm 150^\circ C}$ can raise the temperature of the container+water to ${\rm 45^\circ C}$.  The correct answer is (d)

Author: Ali Nemati
Page Published: 4/17/2021