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A 2.8-kg piece of aluminum at 28.5°C is placed in 1.0 kg of water in a Styrofoam container at room temperature (20.0°C). Estimate the net change in entropy of the system.

Short Answer

Expert verified

The net increase of entropy is \(0.64\;{\rm{J/K}}\).

Step by step solution

01

Concepts

The lost heat by aluminum is equal to the absorbed heat by water.

The change in entropy is\(\Delta S = \frac{Q}{T}\)

02

Given data

The mass of water is \({m_1} = 1.0\;{\rm{kg}}\).

The mass of aluminum is \({m_2} = 2.8\;{\rm{kg}}\).

The initial temperature of the aluminum is \({T_2} = {28.5 \circ }{\rm{C}}\).

The initial temperature of the water is \({T_1} = {20.0 \circ }{\rm{C}}\)

Let T be the final temperature of the system.

03

Calculation

The specific heat of water is \({c_1} = 4186\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}\) , and the specific heat of aluminum is \({c_2} = 900\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}\).

Now for thermal equilibrium you get,

\(\begin{array}{c}{m_1}{c_1}\left( {T - {T_1}} \right) = {m_2}{c_2}\left( {{T_2} - T} \right)\\{m_1}{c_1}T - {m_1}{c_1}{T_1} = {m_2}{c_2}{T_2} - {m_2}{c_2}T\\\left( {{m_1}{c_1} + {m_2}{c_2}} \right)T = {m_2}{c_2}{T_2} + {m_1}{c_1}{T_1}\\T = \frac{{{m_2}{c_2}{T_2} + {m_1}{c_1}{T_1}}}{{{m_1}{c_1} + {m_2}{c_2}}}\end{array}\)

Now substituting the values in the above equation, you get,

\(\begin{array}{c}T = \frac{{\left[ {\left( {2.8\;{\rm{kg}}} \right)\left( {900\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}} \right)\left( {{{28.5} \circ }{\rm{C}}} \right)} \right] + \left[ {\left( {1.0\;{\rm{kg}}} \right)\left( {4186\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}} \right)\left( {{{20} \circ }{\rm{C}}} \right)} \right]}}{{\left[ {\left( {1.0\;{\rm{kg}}} \right)\left( {4186\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}} \right)} \right] + \left[ {\left( {2.8\;{\rm{kg}}} \right)\left( {900\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}} \right)} \right]}}\\ = {23.19 \circ }{\rm{C}}\end{array}\)

The amount of heat absorbed by the water equals the heat released by the aluminum.

The heat lost by aluminum is,

\(\begin{array}{c}Q = {m_1}{c_1}\left( {T - {T_1}} \right)\\ = \left( {1.0\;{\rm{kg}}} \right)\left( {4186\;{\rm{J/kg}}{ \cdot \circ }{\rm{C}}} \right)\left( {{{23.19} \circ }{\rm{C}} - {{20} \circ }{\rm{C}}} \right)\\ = 13353\;{\rm{J}}\end{array}\)

Now the average temperature of the water is,

\(\begin{array}{c}{{T'}_1} = \frac{{{T_1} + T}}{2}\\ = \frac{{{{20} \circ }{\rm{C}} + {{23.19} \circ }{\rm{C}}}}{2}\\ = {21.60 \circ }{\rm{C}}\\ = 294.60\;{\rm{K}}\end{array}\)

Now the average temperature of the aluminum is,

\(\begin{array}{c}{{T'}_2} = \frac{{{T_2} + T}}{2}\\ = \frac{{{{28.5} \circ }{\rm{C}} + {{23.19} \circ }{\rm{C}}}}{2}\\ = {25.85 \circ }{\rm{C}}\\ = 298.85\;{\rm{K}}\end{array}\)

Now the rate of change in entropy for the hot water is \(\Delta {S_1} = - \frac{Q}{{{{T'}_1}}}\).

The rate of change in entropy for the cold water is \(\Delta {S_2} = \frac{Q}{{{{T'}_2}}}\).

Therefore the rate of change of net entropy is,

\(\begin{array}{c}\Delta S = \Delta {S_1} + \Delta {S_2}\\ = + \frac{Q}{{{{T'}_1}}} - \frac{Q}{{{T_2}}}\\ = Q\left( {\frac{1}{{{{T'}_1}}} - \frac{1}{{{{T'}_2}}}} \right)\end{array}\)

Now substituting the values in the above equation, you get,

\(\begin{array}{c}\Delta S = Q\left( {\frac{1}{{{{T'}_2}}} - \frac{1}{{{{T'}_1}}}} \right)\\ = 13353\;{\rm{J}} \times \left( {\frac{1}{{294.60\;{\rm{K}}}} - \frac{1}{{298.85\;{\rm{K}}}}} \right)\\ = 0.64\;{\rm{J/K}}\end{array}\)

Hence, the net increase of entropy is \(0.64\;{\rm{J/K}}\).

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Most popular questions from this chapter

(II) An inventor claims to have built an engine that produces 2.00 MW of usable work while taking in 3.00 MW of thermal energy at 425 K, and rejecting 1.00 MW of thermal energy at 215 K. Is there anything fishy about his claim? Explain.

Two 1100 kg cars are traveling at a speed of\({\bf{85 km/h}}\)in opposite directions when they collide and are brought to rest. Estimate the change in entropy of the universe as a result of this collision. Assume\(T = {\bf{20\circ C}}\).

Question: An ideal heat pump is used to maintain the inside temperature of a house at \({T_{{\rm{in}}}} = 22{\rm{^\circ C}}\) when the outside temperature is \({T_{{\rm{out}}}}\). Assume that when it is operating, the heat pump does work at a rate of 1500 W. Also assume that the house loses heat via conduction through its walls and other surfaces at a rate given by \(\left( {650\;{{\rm{W}} \mathord{\left/

{\vphantom {{\rm{W}} {{\rm{^\circ C}}}}} \right.} {{\rm{^\circ C}}}}} \right)\left( {{T_{{\rm{in}}}} - {T_{{\rm{out}}}}} \right)\). (a) For what outside temperature would the heat pump have to operate all the time in order to maintain the house at an inside temperature of 22°C? (b) If the outside temperature is 8°C, what percentage of the time does the heat pump have to operate in order to maintain the house at an inside temperature of 22°C?

Question: (II) A 1.0-L volume of air initially at 3.5 atm of (gauge) pressure is allowed to expand isothermally until the pressure is 1.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a PV diagram, including numbers and labels for the axes.

What is the change in entropy of\({\bf{1}}{\bf{.00}}\;{{\bf{m}}{\bf{3}}}\)water at 0°C when it is frozen to ice at 0°C?

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