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Expert-verified Found in: Page 503 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # (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.

1. The amount of heat required is $$619200{\rm{ J}}$$.
2. The time taken for the stages are $$0.418{\rm{ s}}$$, $$3.34{\rm{ s}}$$, $$4.19{\rm{ s}}$$, $$22.56{\rm{ s}}$$, $$0.456{\rm{ s}}$$.
3. The graph is constructed by plotting calculated time periods on the x-axis and temperature on the y-axis.
See the step by step solution

## Step 1:Calculate the amount of heat transfer.

(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 $${\rm{619200 J}}$$.

## Step 2: Arrive at an expression to calculate the force of cardiac contraction

(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.

## Step 3: Plot the graph of temperature versus time

(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. ### Want to see more solutions like these? 