Question

How much energy is needed to convert $23.0{\text{ grams}}$ of ice at $- {10.0^ \circ }{\text{C}}$ into steam at ${109^ \circ }{\text{C}}$?

Answer 1

To solve this we must know the equation to calculate the energy required during phase change and temperature change. In this process when $23.0{\text{ grams}}$ of ice at $- {10.0^ \circ }{\text{C}}$ is converted into steam at ${109^ \circ }{\text{C}}$ there are five steps involved.

Complete solution:
We are given that $23.0{\text{ grams}}$ of ice at $- {10.0^ \circ }{\text{C}}$ is converted into steam at ${109^ \circ }{\text{C}}$. In this process, there are five steps involved.
Heating ice from ${0^ \circ }{\text{C}}$ to $- {10.0^ \circ }{\text{C}}$: Here, temperature change occurs. Thus,
${Q_1} = mC\Delta T$
Where ${Q_1}$ is the energy required,
$m$ is the mass,
$C$ is the specific heat capacity of ice,
$\Delta T$ is the change in temperature.
Substitute $23.0{\text{ grams}} = 23.0 \times {10^{ - 3}}{\text{ kg}}$ for the mass, $2.108{\text{ kJ/kg}}{ \cdot ^ \circ }{\text{C}}$ for the specific heat capacity of ice, $\left( {0 - \left( { - 10} \right)} \right) = {10.0^ \circ }{\text{C}}$ for the change in temperature. Thus,
${Q_1} = 23.0 \times {10^{ - 3}}{\text{ kg}} \times 2.108{\text{ kJ/kg}}{ \cdot ^ \circ }{\text{C}} \times {10.0^ \circ }{\text{C}}$
${Q_1} = 0.48484{\text{ kJ}}$
Melting ice at ${0^ \circ }{\text{C}}$: Here, phase change occurs. Thus,
${Q_2} = m\Delta H$
Where ${Q_2}$ is the energy required,
$m$ is the mass,
$\Delta H$ is the latent heat of melting.
Substitute $23.0{\text{ grams}} = 23.0 \times {10^{ - 3}}{\text{ kg}}$ for the mass, $334{\text{ kJ/kg}}$ for the latent heat of melting. Thus,
${Q_2} = 23.0 \times {10^{ - 3}}{\text{ kg}} \times 334{\text{ kJ/kg}}$
${Q_2} = 7.682{\text{ kJ}}$
Heating liquid water from ${0^ \circ }{\text{C}}$ to ${100^ \circ }{\text{C}}$: Here, temperature change occurs. Thus,
${Q_3} = mC\Delta T$
Where ${Q_3}$ is the energy required,
$m$ is the mass,
$C$ is the specific heat capacity of water,
$\Delta T$ is the change in temperature.
Substitute $23.0{\text{ grams}} = 23.0 \times {10^{ - 3}}{\text{ kg}}$ for the mass, $4.186{\text{ kJ/kg}}{ \cdot ^ \circ }{\text{C}}$ for the specific heat capacity of water, $\left( {100 - 0} \right) = {100^ \circ }{\text{C}}$ for the change in temperature. Thus,
${Q_3} = 23.0 \times {10^{ - 3}}{\text{ kg}} \times 4.186{\text{ kJ/kg}}{ \cdot ^ \circ }{\text{C}} \times {100^ \circ }{\text{C}}$
${Q_3} = 9.6278{\text{ kJ}}$
Evaporating liquid water at ${100^ \circ }{\text{C}}$: Here, phase change occurs. Thus,
${Q_4} = m\Delta H$
Where ${Q_4}$ is the energy required,
$m$ is the mass,
$\Delta H$ is the latent heat of evaporation.
Substitute $23.0{\text{ grams}} = 23.0 \times {10^{ - 3}}{\text{ kg}}$ for the mass, $2256{\text{ kJ/kg}}$ for the latent heat of evaporation. Thus,
${Q_4} = 23.0 \times {10^{ - 3}}{\text{ kg}} \times 2256{\text{ kJ/kg}}$
${Q_4} = 51.888{\text{ kJ}}$
Heating steam from ${100^ \circ }{\text{C}}$ to ${109^ \circ }{\text{C}}$: Here, temperature change occurs. Thus,
${Q_5} = mC\Delta T$
Where ${Q_5}$ is the energy required,
$m$ is the mass,
$C$ is the specific heat capacity of steam,
$\Delta T$ is the change in temperature.
Substitute $23.0{\text{ grams}} = 23.0 \times {10^{ - 3}}{\text{ kg}}$ for the mass, $1.996{\text{ kJ/kg}}{ \cdot ^ \circ }{\text{C}}$ for the specific heat capacity of steam, $\left( {109 - 100} \right) = {9^ \circ }{\text{C}}$ for the change in temperature. Thus,
${Q_5} = 23.0 \times {10^{ - 3}}{\text{ kg}} \times 1.996{\text{ kJ/kg}}{ \cdot ^ \circ }{\text{C}} \times {9^ \circ }{\text{C}}$
${Q_5} = 0.413172{\text{ kJ}}$
Now, the total energy required is the summation of the energies required at each step. Thus,
$Q = {Q_1} + {Q_2} + {Q_3} + {Q_4} + {Q_5}$
$Q = \left( {0.48484 + 7.682 + 9.6278 + 51.888 + 0.413172} \right){\text{ kJ}}$
$Q = 70.04{\text{ kJ}}$
Thus, the energy needed to convert $23.0{\text{ grams}}$ of ice at $- {10.0^ \circ }{\text{C}}$ into steam at ${109^ \circ }{\text{C}}$ is $70.04{\text{ kJ}}$.

Note: Remember the steps to calculate the energy needed. The steps are as follows:
1-Heating ice from ${0^ \circ }{\text{C}}$ to $- {10.0^ \circ }{\text{C}}$.
2-Melting ice at ${0^ \circ }{\text{C}}$.
3-Heating liquid water from ${0^ \circ }{\text{C}}$ to ${100^ \circ }{\text{C}}$.
4-Evaporating liquid water at ${100^ \circ }{\text{C}}$.
5-Heating steam from ${100^ \circ }{\text{C}}$ to ${109^ \circ }{\text{C}}$.