Chapter 22, Problem 12PEB

If the insolation of the Sun shining on concrete is 7.3 10² W/m², what is the change in temperature of a 1.5 m² by 5.0 cm thick layer of concrete in 1 hr? (Assume the albedo of the concrete is 0.55, the specific heat of concrete is 0.16 cal/gC°, and the density of concrete is 2.4 g/cm³.)

To determine

The change in temperature of a $1.5{\text{ m}}^2$ by $5.0\text{ cm}$ thick layer of concrete in $1\text{ hr}$.

Answer to Problem 12PEB

Solution:

$14.76°\text{C}$

Explanation of Solution

Given data:

Insolation of the sun is $7.3\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$.

The albedo of the concrete is $0.55$.

The area of the layer $1.5{\text{ m}}^2$.

Time is 1 hr.

Density of concrete is $2.4\frac{\text{g}}{{\text{cm}}^{\text{2}}}$.

Thickness of the layer is $5\text{ cm}$.

Specific heat of concrete is $0.16\frac{\text{cal}}{\text{gC}°}$.

Formula used:

Write the equation for the incoming solar radiation by the reflected solar radiation to determine the albedo.

$\alpha =\frac{\text{reflected solar radiation}}{\text{Insolation}}$

Here, $\alpha$ is the insolation.

Write the equation for energy from the absorbed solar radiation:

$\text{absorbed energy}=At\left(\text{insolation}-\text{reflected solar radiation}\right)$

Here, $A$ is the area of the layer and $t$ is time.

Write the formula for density.

$\rho =\frac{m}{V}$

Here, $m$ is the mass and $V$ is the volume.

Write the formula for volume.

$V=zA$

Here, $z$ is the thickness.

Write the formula for heat.

$Q=mc\Delta T$

Here, $m$ is the mass, $c$ is the specific heat and $\Delta T$ is the change in temperature.

Explanation:

Determine reflected radiation:

Recall the equation for the incoming solar radiation by the reflected solar radiation to determine the albedo, denoted by alpha.

$\alpha =\frac{\text{reflected solar radiation}}{\text{incoming solar radiation}}$

Substitute $7.3\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$ for $\alpha$ and $0.55$ for Insolation.

$\begin{array}{c}7.3\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}=\frac{\text{reflected solar radiation}}{0.55}\\ \text{reflected solar radiation}=\left(7.3\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\right)\left(0.55\right)\\ =4\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\end{array}$

Convert hours to seconds:

$\begin{array}{c}1\text{ hr}=1\text{ hr}\left(\frac{60\text{ min}}{1\text{ hr}}\right)\left(\frac{60\text{ sec}}{1\text{ min}}\right)\\ =3600\text{ sec}\end{array}$

Determine energy:

Recall the equation for energy from the absorbed solar radiation:

$\text{absorbed energy}=At\left(\text{insolation}-\text{reflected solar radiation}\right)$

Substitute $1.5{\text{ m}}^2$ for $A$, $3600\text{ sec}$ for $t$, $7.3\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$ for insolation and $4\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$ for reflected solar radiation:

$\begin{array}{c}\text{absorbed energy}=\left(1.5{\text{ m}}^2\right)\left(3600\text{ sec}\right)\left(7.3\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}-4\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\right)\\ =\left(5400\right)\left(330\right){\text{ m}}^2\cdot \text{sec}\cdot \frac{\frac{\text{J}}{\text{sec}}}{{\text{m}}^{\text{2}}}\\ =1.78\times {10}^6\text{ J}\end{array}$

Determine the mass of concrete:

Convert ${\text{m}}^{\text{2}}$ to ${\text{cm}}^{\text{2}}$.

$\begin{array}{c}1.5{\text{ m}}^2=1.5{\text{ m}}^2\left(\frac{{10}^4\text{ cm}}{1{\text{ m}}^2}\right)\\ =1.5\times {10}^4\text{ cm}\end{array}$

Recall the formula for volume.

$V=zA$

Recall the formula for density.

$\rho =\frac{m}{V}$

Substitute $zA$ for $V$.

$\begin{array}{c}\rho =\frac{m}{zA}\\ m=\rho zA\end{array}$

Substitute $2.4\frac{\text{g}}{{\text{cm}}^{\text{2}}}$ for $\rho$, $5\text{ cm}$ for $z$ and $1.5\times {10}^4\text{ cm}$ for $A$.

$\begin{array}{c}m=\left(2.4\frac{\text{g}}{{\text{cm}}^{\text{2}}}\right)\left(5\text{ cm}\right)\left(1.5\times {10}^4\text{ cm}\right)\\ =1.8\times {10}^5\text{ g}\end{array}$

Convert energy to calorie heat:

$\begin{array}{c}1.78\times {10}^6\text{ J}=1.78\times {10}^6\text{ J}\left(\frac{1\text{ cal}}{\text{4}\text{.184 J}}\right)\\ =4.25\times {10}^5\text{ cal}\end{array}$

Determine the temperature change of concrete.

Recall the formula for heat:

$Q=mc\Delta T$

Substitute $0.16\frac{\text{cal}}{\text{gC}°}$ for $c$, $4.25\times {10}^5\text{ cal}$ for $Q$ and $1.8\times {10}^5\text{ g}$ for $m$.

$\begin{array}{c}\left(4.25\times {10}^5\text{ cal}\right)=\left(1.8\times {10}^5\text{ g}\right)\left(0.16\frac{\text{cal}}{\text{gC}°}\right)\Delta T\\ \Delta T=\frac{\left(4.25\times {10}^5\text{ cal}\right)}{\left(1.8\times {10}^5\text{ g}\right)\left(0.16\frac{\text{cal}}{\text{gC}°}\right)}\\ =14.76°\text{C}\end{array}$

Conclusion:

The change in temperature is $14.76°\text{C}$.