Question

How much heat (in $kJ$ ) is required to convert $423g$ of liquid ${{H}_{2}}O$ at $24.0{}^\circ C$ into steam at $152{}^\circ C?$ (Assume that the specific heat of liquid water is $4.184\text{ }J/g{}^\circ C,$ the specific heat of steam is $2.078\text{ }J/g{}^\circ C,$ and that both values are constant over the given temperature ranges?

Answer 1

The Heat of Vaporization is the amount of heat that needs to be absorbed at a constant temperature to vaporize a specific amount of liquid. The kinetic energy of the steam is demonstrated to be higher than the kinetic energy of the fluid if the solutions of vapour and liquid states are compared. We will be using equation $Q=mcDe$<$T.$

Complete answer:
$Q$ is energy lost or gained, m is mass of water, $c$ is specific heat, of which there will be two; one for water and one for steam, and $\Delta T$ which is final temperature minus initial temperature.
There are four steps required to answer this question. First, we will determine $Q$ of liquid water. Second we will determine $Q$ required to convert given mass of liquid water to steam. n we will determine $Q$ for steam. Then we will add all values for $Q$ to get the total amount of energy needed to convert a given mass of liquid water to steam. value of $Q$ will be in Joules, so we will have to convert $J$ to $kJ$ by dividing $J$ by $1000$
We have, $m=423g\left( {{H}_{2}}O \right)$ ; ${{c}_{{{H}_{2}}{{O}_{(l)}}}}=4.185J/{{g}^{o}}C$ and ${{c}_{{{H}_{2}}{{O}_{(g)}}}}=2.078J/{{g}^{o}}C$
Also we have, $\Delta {{H}_{v}}=2257\text{ }J/g$ ; $\Delta {{T}_{{{H}_{2}}O}}_{_{(l)}}=({{100.0}^{\circ }}C-{{24.0}^{\circ }}C)={{76.0}^{\circ }}C$ and $\Delta {{T}_{{{H}_{2}}{{O}_{(g)}}}}=({{152}^{\circ }}C-{{100.0}^{\circ }}C)={{52.0}^{\circ }}C$
STEP 1: We have to determine $Q$ for liquid water;
$Q=\left( m \right)\times \left( {{c}_{{{H}_{2}}{{O}_{(l)}}}} \right)\times \left( \Delta {{T}_{{{H}_{2}}O}}_{_{(l)}} \right)$
Substituting the given values;
$Q=\left( 423g \right)\times \left( 4.185J/{{g}^{o}}C \right)\times \left( {{76}^{o}}C \right)$
$\Rightarrow Q=24166.78J$
STEP 2: Now we have to determine $Q$ for phase change from liquid to steam;
$Q=\left( m \right)\times \left( \Delta {{H}_{v}} \right)$
Substituting the given values;
$Q=\left( 423g \right)\times (2257J/g)$
$\Rightarrow Q=974711J$
STEP 3: Now we have to determine $Q$ for temperature change from ${{100.0}^{\circ }}C~$ to ${{152}^{\circ }}C$ ;
$Q=\left( m \right)\times \left( {{c}_{{{H}_{2}}{{O}_{(g)}}}} \right)\times \left( \Delta {{T}_{{{H}_{2}}{{O}_{(g)}}}} \right)$
Substituting the given values;
$Q=\left( 423g \right)\times (2.078J/{{g}^{o}}C)\times \left( {{52}^{o}}C \right)$
$\Rightarrow Q=45707.66J$
Therefore, by adding all the values of $Q$ so that we get ${{Q}_{total}}$ ;
${{Q}_{total}}=24166.784J~+~974711J~+~45707.668J=1044585J$
$\Rightarrow {{Q}_{total}}=1040.000J$
Now, we have convert joules into kilojoules;
$1kJ=1000J$
$1040000J\times \dfrac{1kJ}{1000J}=1040kJ$

Note:
This is a very simple problem which is directly based on a single formula. Students should remember the value of heat of vaporization of water or else they would never be able to come to the conclusion of the given problem. In other words, we can say that vaporization heat is the total quantity of heat required to convert a specific quantity of liquid into its vapor form without any major temperature increase.