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Determine the heat flow in 15 min15 \mathrm{~min} through a 0.10cm0.10-\mathrm{cm}-thick copper plate with crosssectional area 150 cm2150 \mathrm{~cm}^2 if the temperature of the hot side is 990°C990{\degree} \mathrm{C} and the temperature of the cool side is 5°C5{\degree} \mathrm{C}.

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Answered 2 years ago
Answered 2 years ago
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Given\textbf{Given}

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

A=150 cm2A=150 \text{ cm}^2

T1=5CT_1= 5 ^\circ\text{C}

T2=990CT_2= 990 ^\circ\text{C}

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

Kcopper=220 JmC 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=KAt(T2T1)L=KAtΔTLQ=\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, T1T_1 outdoor temperature, T2T_2 indoor temperature, t time and K thermal conductivity.

First we need to convert surface area using 1 m=100 cm1 \text{ m}= 100 \text{ cm} from cm2^2 to m2^2

A=0.015 m2A= 0.015 \text{ m}^2

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

Q=2200.015900(9905)0.001Q=3×109 J\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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