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A pitcher with 1 mm thick porous walls contains 10 kg of water. Water comes to its outer surface...

Question:

A pitcher with 1 mm thick porous walls contains 10 kg of water. Water comes to its outer surface and evaporates at a rate of 0.1 g/s{eq}^{-1} {/eq}. The surface area of the pitcher (one side) is 200 cm{eq}^2 {/eq}. The room temperature = 42 degrees Celsius, latent heat of vaporization = {eq}2.27 \times 10^6 {/eq} J/kg{eq}^{-1} {/eq}, and the thermal conductivity of the porous walls = 0.80 J/s{eq}^{-1} {/eq}m{eq}^{-1} {/eq} degrees Celsius{eq}^{-1} {/eq}. Calculate the temperature of water in the pitcher when it attains a constant value.

Heat Transfer:

When the temperature of a system increases and decreases, then the heat transfer from the system may occur. Due to heat transfer, either the system may become hotter or colder. Generally, the heat transfer is denoted by Joule in the International Standard of the unit system.

Answer and Explanation:


Given

  • The thickness is t = 1 mm = 0.001 m
  • The mass of the contained water is m = 10 kg
  • The rate of evaporation is e = 0.1 g/s
  • The surface area of the pitcher is {eq}A = 200\;{\rm{c}}{{\rm{m}}^{\rm{2}}} = 0.02\;{{\rm{m}}^{\rm{2}}} {/eq}
  • The room temperature is {eq}T = 42{\rm{^\circ C}} {/eq}
  • The latent heat of vaporization is {eq}L = 2.27 \times {10^6}\;{\rm{J/kg}} {/eq}
  • The thermal conductivity of the porous wall is {eq}k = 0.8\;{\rm{J/m^\circ C}} \cdot {\rm{s}} {/eq}


The change in heat transfer is calculated as,

{eq}\begin{align*} dQ &= L \times m\\ &= \left( {2.27 \times {{10}^6}\;{\rm{J/kg}}} \right) \times \left( {10\;{\rm{kg}}} \right)\\ &= 2.27 \times {10^7}\;{\rm{J}} \end{align*} {/eq}


The change in time is calculated as,

{eq}\begin{align*} e = 0.1\;{\rm{g/s}}\\ \dfrac{m}{t} = \dfrac{{0.1\;{\rm{g}}}}{{1\;{\rm{s}}}}\\ m = 0.1\;g,\;then\;t = 1\;{\rm{s}}\\ {\rm{So,}}\;{\rm{if}}\;m = 1\;{\rm{g,}}\;{\rm{then}}\;t = 10\;{\rm{s}}\\ {\rm{For}}\;m = 1000\;{\rm{g,}}\;{\rm{then}}\;t = 10000\;{\rm{s}}\\ m = 1\;{\rm{kg,}}\;{\rm{then}}\;t = 10000\;{\rm{s}}\\ {\rm{Then}}\;{\rm{For,}}\;m = 10\;{\rm{kg}}\\ dt = 100000\;{\rm{s}}\\ dt = {10^5}\;{\rm{s}} \end{align*} {/eq}


The rate of heat transfer is calculated as,

{eq}\begin{align*} \dfrac{{dQ}}{{dt}} &= \dfrac{{2.27 \times {{10}^7}\;{\rm{J}}}}{{{{10}^5}\;{\rm{s}}}}\\ &= 2.27 \times {10^2}\;{\rm{J/s}} \end{align*} {/eq}


The temperature of water in the pitcher when it attains a constant value is calculated as,

{eq}\begin{align*} \dfrac{{dQ}}{{dt}} &= \dfrac{{\Delta T \times k \times A}}{t}\\ \dfrac{{dQ}}{{dt}} &= \dfrac{{\left( {T - {T_1}} \right) \times k \times A}}{t}\\ \left( {2.27 \times {{10}^2}\;{\rm{J/s}}} \right) &= \dfrac{{\left( {42{\rm{^\circ C}} - {T_1}} \right) \times \left( {0.8\;{\rm{J/m^\circ C}} \cdot {\rm{s}}} \right) \times \left( {0.02\;{{\rm{m}}^{\rm{2}}}} \right)}}{{\left( {0.001\;{\rm{m}}} \right)}}\\ \left( {42{\rm{^\circ C}} - {T_1}} \right) &= \dfrac{{\left( {2.27 \times {{10}^2}\;{\rm{J/s}}} \right) \times \left( {0.001\;{\rm{m}}} \right)}}{{\left( {0.8\;{\rm{J/m^\circ C}} \cdot {\rm{s}}} \right) \times \left( {0.02\;{{\rm{m}}^{\rm{2}}}} \right)}}\\ {T_1} &= 42{\rm{^\circ C}} - \dfrac{{\left( {2.27 \times {{10}^2}\;{\rm{J/s}}} \right) \times \left( {0.001\;{\rm{m}}} \right)}}{{\left( {0.8\;{\rm{J/m^\circ C}} \cdot {\rm{s}}} \right) \times \left( {0.02\;{{\rm{m}}^{\rm{2}}}} \right)}}\\ {T_1} &= 42{\rm{^\circ C}} - 14.188{\rm{^\circ C}}\\ {T_1} &= 27.812{\rm{^\circ C}} \end{align*} {/eq}


Thus, the temperature of water in the pitcher when it attains a constant value is {eq}27.812{\rm{^\circ C}} {/eq}


Learn more about this topic:

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Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

Chapter 17 / Lesson 12
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