Consider a 1250-W iron whose base plate is made of 0.6-cm-thick aluminum alloy 2024-T6 (rho =...

Question:

Consider a 1250-W iron whose base plate is made of 0.6-cm-thick aluminum alloy 2024-T6 ({eq}\rho {/eq} = 2770 kg/m{eq}^3 {/eq} and c{eq}\rho {/eq}= 875 J/kg {eq}^\circ {/eq}C). The base plate has a surface area of 0.04 m{eq}^2 {/eq}. Initially, the iron is in thermal equilibrium with the ambient air at 22{eq}^\circ {/eq}C. Assuming 90 percent of the heat generated in the resistance wires is transferred to the plate, determine the minimum time needed for the plate temperature to reach 206{eq}^\circ {/eq}C.

Heat transfer

the basic cause of heat transfer is the difference in temperature.when two bodies are at different temperatures and connect to each other than the heat is transfer to a higher temperature body to lower the temperature body until both body come in equilibrium condition.

Answer and Explanation:

Given

{eq}\begin{align*} A &= 0.04\;{{\rm{m}}^{\rm{2}}}\\ L &= 0.6\;m\\ \rho &= 2770\;{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\\ {c_p} &= 875\;{\rm{j/kg^\circ C}} \end{align*} {/eq}

{eq}\begin{align*} {\eta _{heat\;generation}} &= 0.9\\ {T_\infty } &= 22\;{\rm{^\circ C}}\\ h &= 12\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}{\rm{.k}}\\ {T_{plate}} &= \left( {\dfrac{{206 + 22}}{2}} \right) \end{align*}{/eq}

{eq}{T_{plate}} = 114\;{\rm{^\circ C}}{/eq}

First we determine the mass of iron base plate;

{eq}\begin{align*} m &= \rho .V\\ &= \rho LA\\ &= 2770 \times 0.6 \times {10^{ - 2}} \times 0.04\\ &= 0.6648\;{\rm{kg}} \end{align*}{/eq}

{eq}\begin{align*} {Q_{in}} &= 0.9 \times 1250\\ &= 1187.5\;{\rm{W}} \end{align*}{/eq}

Now determine heat loss;

{eq}\begin{align*} {Q_{loss}} &= hA\left( {{T_{plate}} - {T_\infty }} \right)\\ &= 12 \times 0.04 \times \left( {114 - 22} \right)\\ &= 44.16\;{\rm{W}} \end{align*}{/eq}

By energy equation;

{eq}\begin{align*} {E_{in}} - {E_{out}} &= m{c_p}\Delta T\\ {Q_{in}}.\Delta t - {Q_{out}}.\Delta t &= m{c_p}\Delta T\\ \Delta t &= \dfrac{{m{c_p}\Delta T}}{{{Q_{in}} - {Q_{out}}}}\\ &= \dfrac{{0.648 \times 875 \times \left( {206 - 22} \right)}}{{1187.5 - 44.16}} \end{align*}{/eq}

{eq}\Delta t = 91.24\;{\rm{sec}}{/eq}


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Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

Chapter 17 / Lesson 12
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