Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \(0.64\;{\rm{J/K}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free