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# Determine the heat flow in $15 \mathrm{~min}$ through a $0.10-\mathrm{cm}$-thick copper plate with crosssectional area $150 \mathrm{~cm}^2$ if the temperature of the hot side is $990{\degree} \mathrm{C}$ and the temperature of the cool side is $5{\degree} \mathrm{C}$.

Solution

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$\textbf{Given}$

$L=0.1 \text{ cm}= 0.001\text{ m}$

$A=150 \text{ cm}^2$

$T_1= 5 ^\circ\text{C}$

$T_2= 990 ^\circ\text{C}$

$t=15 \text{ min}=900 \text{ s}$

$K_{copper}=220 \text{ } \dfrac{\text{J}}{\text{m} ^\circ\text{C s }}$

The amount of heat conducted through the steel is given by the formula:

$Q=\dfrac{K \cdot A \cdot t (T_2 - T_1)}{L}=\dfrac{K \cdot A \cdot t \cdot \Delta T}{L}$

where L is thickness, A surface, $T_1$ outdoor temperature, $T_2$ indoor temperature, t time and K thermal conductivity.

First we need to convert surface area using $1 \text{ m}= 100 \text{ cm}$ from cm$^2$ to m$^2$

$A= 0.015 \text{ m}^2$

By including the given data and the calculated surface in the heat formula we get:

\begin{align*} Q&=\dfrac{220 \cdot 0.015 \cdot 900 \cdot (990 - 5) }{0.001}\\ &\boxed{Q= 3\times 10^{9} \text{ J}} \end{align*}

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