As a coffee enthusiast, you wish to keep your coffee as hot as possible for as long as possible....

Question:

As a coffee enthusiast, you wish to keep your coffee as hot as possible for as long as possible. You wish to find the rate of heat loss of a typical coffee mug. Assume the mug to be a cylinder which has a lid made out of the same material as the rest of the cylinder, and the thickness of the walls is 1.00 cm.

If the coffee in the mug is 79.2{eq}^{\circ} {/eq}C and the room temperature is 23.2{eq}^{\circ} {/eq}C, how much more heat doe the mug lose through conduction than from radiated energy?

Heat Transfer:

It is the process in which heat is carried due to the varying temperatures of two surfaces or bodies. Naturally, the transfer of thermal energy takes place from high temperature to low temperature.

Answer and Explanation:


Given Data


  • Thickness of the wall is, {eq}d = 1.00\;{\rm{cm}} {/eq}.
  • Temperature of coffee in mug is, {eq}T = 79.2^\circ {\rm{C}} {/eq}.
  • Temperature of room is, {eq}{T_C} = 23.2^\circ {\rm{C}} {/eq}.


The expression for heat transfer through conduction is,

{eq}\begin{align*} {{\dot Q}_{cond}} &= \dfrac{Q}{t}\\ {{\dot Q}_{cond}} &= \dfrac{{kA\Delta T}}{d}\\ \dfrac{{{{\dot Q}_{cond}}}}{A} &= \dfrac{{k\Delta T}}{d} \end{align*} {/eq}


Here k is the thermal conductivity of all sides of mug whose value is {eq}0.873\;{\rm{W/m}} \cdot ^\circ {\rm{C}} {/eq}.

And {eq}\Delta T = T - {T_C} {/eq} is the change in temperature.


Substituting the values in above expression.

{eq}\begin{align*} \dfrac{{{{\dot Q}_{cond}}}}{A} &= \dfrac{{0.873 \times 56}}{{0.01}}\\ \dfrac{{{{\dot Q}_{cond}}}}{A} &= 4888.8\;{\rm{W/}}{{\rm{m}}^2} \end{align*} {/eq}


The expression for heat transfer through radiation is,

{eq}\dfrac{{{Q_{radiation}}}}{A} = \sigma \varepsilon \left( {{T^4} - T_C^4} \right) {/eq}


Here, {eq}\sigma {/eq} is the Stefan-Boltzmann constant whose value is {eq}5.67 \times {10^{ - 8}}\;{\rm{W/}}{{\rm{m}}^2} {/eq} and {eq}\varepsilon {/eq} is the emissivity of mug whose value is 0.715.


Substituting the values in above expression.

{eq}\begin{align*} \dfrac{{{Q_{radiation}}}}{A} &= 0.715 \times 5.67 \times {10^{ - 8}}\left( {{{\left( {79.2 + 273} \right)}^4} - {{\left( {23.2 + 273} \right)}^4}} \right)\\ \dfrac{{{Q_{radiation}}}}{A} &= 311.74\;{\rm{W/}}{{\rm{m}}^2} \end{align*} {/eq}


The ratio is given by,

{eq}\begin{align*} \dfrac{{{{\dot Q}_{cond}}}}{{{Q_{radiation}}}} &= \dfrac{{4888.8}}{{311.74}}\\ \dfrac{{{{\dot Q}_{cond}}}}{{{Q_{radiation}}}} &= 15.68 \end{align*} {/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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