(a) How much heat transfer is necessary to raise the temperature of a\({\rm{0}}{\rm{.200 kg}}\) piece of ice from\({\rm{ - 2}}{{\rm{0}}^{\rm{o}}}{\rm{C}}\)to \({\rm{ - 130}}{{\rm{}}^{\rm{o}}}{\rm{C}}\) , including the energy needed for phase changes?

(b) How much time is required for each stage, assuming a constant\({\rm{20}}{\rm{.0 kJ/s}}\)rate of heat transfer?

(c) Make a graph of temperature versus time for this process.

(a)

The phase of a substance changes with temperature. The ice becomes water when it reached a temperature above\({{\rm{0}}^{\rm{o}}}{\rm{C}}\). At\({\rm{10}}{{\rm{0}}^{\rm{o}}}{\rm{C}}\), this water boils and turns to steam. The amount of heat transfer required to raise the temperature is given as follows:

\({\rm{Q = mc\Delta T}}\)

Here,\({\rm{Q}}\)is the amount of heat transfer,\({\rm{m}}\)is the mass of the substance,\({\rm{c}}\)is the specific heat capacity of the substance and\({\rm{\Delta T}}\)is the change in temperature. The mass of the ice is given, and the specific heat capacities of different forms of ice are also known.

\(\begin{aligned} m &= 0.200{\rm{ }}kg\\{c_{ice}} &= 2090{\rm{ }}J/kg \cdot ^\circ C\\{c_{water}} &= 4186{\rm{ }}J/kg \cdot ^\circ C\\{c_{steam}} &= 1520{\rm{ }}J/kg \cdot ^\circ C\end{aligned}\)

The temperature change during rise in temperature in each stage is calculated as follows:

\(\begin{aligned}\Delta {T_{ - 20 - 0}} &= \left( {0^\circ C - 20^\circ C} \right)\\ &= 20^\circ C\\\Delta {T_{0 - 100}} &= \left( {100^\circ C - 0^\circ C} \right)\\ &= 100^\circ C\end{aligned}\)

\(\begin{aligned}\Delta {T_{100 - 130}} &= \left( {130^\circ C - 100^\circ C} \right)\\ &= 30^\circ C\end{aligned}\)

The heat required to change the phase of the substance is calculated as follows:

\(\begin{aligned}Q &= m{L_v}\\Q &= m{L_f}\end{aligned}\)

Here, \({L_f}\) is the latent heat of fusion and \({L_v}\) is the latent heat of vaporization.

\(\begin{aligned}{L_f}{\rm{ }} of {\rm{ }}water &= 334{\rm{ }}kJ/kg\\{L_v}{\rm{ }} of {\rm{ }}water &= 2256{\rm{ }}kJ/kg\end{aligned}\)

The heat transfer required for ice to turn into steam can be calculated as follows.

\(\begin{aligned}Q &= m{c_{ice}}\Delta {T_{ - 20 - 0}} + m{L_f} + m{c_{water}}\Delta {T_{0 - 100}} + m{L_v} + m{c_{steam}}{T_{0 - 100}}\\ &= \left( {0.200{\rm{ }}kg \times 2090{\rm{ }}J/kg \times ^\circ C \times 20{\rm{ }}^\circ C} \right) + \left( {0.200{\rm{ }}kg \times 334 \times {{10}^3}{\rm{ }}J/kg} \right)\\ &+ \left( {0.200{\rm{ }}kg \times 4186{\rm{ }}J/kg \times ^\circ C \times 100{\rm{ }}^\circ C} \right) + \left( {0.200{\rm{ }}kg \times 2256 \times {{10}^3}{\rm{ }}J/kg} \right)\\ &+ \left( {0.200{\rm{ }}kg \times 1520{\rm{ }}J/kg \times ^\circ C \times 30{\rm{ }}^\circ C} \right)\end{aligned}\)

\(\begin{aligned} Q &= 8360{\rm{ }}J + 66800{\rm{ }}J + 83720{\rm{ }}J + 451200{\rm{ }}J + 9120{\rm{ }}J\\ &= 619200{\rm{ }}J\end{aligned}\)

Therefore, amount of heat required is \({\rm{619200 J}}\).

(b)

The rate of heat transfer is the ratio of heat transfer to time. It is related to the time as follows:

\(\begin{aligned}rate &= \frac{Q}{t}\\t &= \frac{Q}{{rate}}\end{aligned}\)

Rate of heat transfer is given as \({\rm{20 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{ J/s}}\). The time taken in each stage is calculated as follows:

\(\begin{aligned}{t_1} &= \frac{{{Q_{m{c_{ice}}\Delta {T_{ - 20 - 0}}}}}}{{20 \times {{10}^3}{\rm{ }}J/s}}\\ &= \frac{{8360{\rm{ }}J}}{{20 \times {{10}^3}{\rm{ }}J/s}}\\ &= 0.418{\rm{ }}s\end{aligned}\)

\(\begin{aligned}{t_2} &= \frac{{{Q_{m{L_f}}}}}{{20 \times {{10}^3}{\rm{ }}J/s}}\\ &= \frac{{66800{\rm{ }}J}}{{20 \times {{10}^3}{\rm{ }}J/s}}\\ &= 3.34{\rm{ }}s\end{aligned}\)

\(\begin{aligned}{t_3} &= \frac{{{Q_{m{c_{water}}\Delta {T_{0 - 100}}}}}}{{20 \times {{10}^3}{\rm{ }}J/s}}\\ &= \frac{{83720{\rm{ }}J}}{{20 \times {{10}^3}{\rm{ }}J/s}}\\ &= 4.19{\rm{ }}s\end{aligned}\)

\(\begin{aligned}{t_4} &= \frac{{{Q_{m{L_v}}}}}{{20 \times {{10}^3}{\rm{ }}J/s}}\\ &= \frac{{451200{\rm{ }}J}}{{20 \times {{10}^3}{\rm{ }}J/s}}\\ &= 22.56{\rm{ }}s\end{aligned}\)

\(\begin{aligned}{t_5} &= \frac{{{Q_{m{c_{steam}}{T_{0 - 100}}}}}}{{20 \times {{10}^3}{\rm{ }}J/s}}\\ &= \frac{{9120{\rm{ }}J}}{{20 \times {{10}^3}{\rm{ }}J/s}}\\ &= 0.456{\rm{ }}s\end{aligned}\)

Thus, the time taken for each stage is calculated.

(c)

Plot the graph by marking the calculated time period on the range of temperature. The graph is plotted as follows:

Thus, the graph is plotted.