Chapter 22, Problem 13PEB

If the insolation of the Sun shining on seawater is 8.7 $\times$ 10² W/m², what is the change in temperature of a 2.0 m² by 10.0 cm thick layer of seawater at the surface in 1 hr? (Assume the albedo of the seawater is 0.4, the specific heat of seawater is 0.92 cal/gC°, and the density of seawater is 1.03 g/cm3.)

To determine

The change in temperature of a $2.0{\text{ m}}^2$ by $10.0\text{ cm}$ thick layer of seawater at the surface in $1\text{ hr}$.

Answer to Problem 13PEB

Solution:

$4.69°\text{C}$

Explanation of Solution

Given data:

Insolation of the sun is $8.7\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$.

The albedo of the sea water is $0.4$.

The area of the layer at the surface $2.0{\text{ m}}^2$.

Time is 1 hr.

Density of seawater is $1.03\frac{\text{g}}{{\text{cm}}^{\text{2}}}$.

Thickness of the layer of seawater is $10\text{ cm}$.

Specific heat of seawater is $0.92\frac{\text{cal}}{\text{gC}°}$.

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 $8.7\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$ for $\alpha$ and $0.4$ for Insolation.

$\begin{array}{c}8.7\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}=\frac{\text{reflected solar radiation}}{0.55}\\ \text{reflected solar radiation}=\left(8.7\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\right)\left(0.4\right)\\ =3.48\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\end{array}$

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 $2.0{\text{ m}}^2$ for $A$, $3600\text{ sec}$ for $t$, $8.7\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$ for insolation and $3.48\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}$ for reflected solar radiation;

$\begin{array}{c}\text{absorbed energy}=\left(2.0{\text{ m}}^2\right)\left(3600\text{ sec}\right)\left(8.7\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}-3.48\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\right)\\ =\left(7200\right)\left(522\right){\text{ m}}^2\cdot \text{sec}\cdot \frac{\frac{\text{J}}{\text{sec}}}{{\text{m}}^{\text{2}}}\\ =3.75\times {10}^6\text{ J}\end{array}$

Determine the mass of seawater:

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

$\begin{array}{c}2.0{\text{ m}}^2=2.0{\text{ m}}^2\left(\frac{{10}^4\text{ cm}}{1{\text{ m}}^2}\right)\\ =2.0\times {10}^4\text{ cm}\end{array}$

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 $1.03\frac{\text{g}}{{\text{cm}}^{\text{2}}}$ for $\rho$, $10\text{ cm}$ for $z$ and $2.0\times {10}^4\text{ cm}$ for $A$.

$\begin{array}{c}m=\left(1.03\frac{\text{g}}{{\text{cm}}^{\text{2}}}\right)\left(10\text{ cm}\right)\left(2.0\times {10}^4\text{ cm}\right)\\ =2.06\times {10}^5\text{ g}\end{array}$

Convert energy to calorie heat:

$\begin{array}{c}3.75\times {10}^6\text{ J}=3.75\times {10}^6\text{ J}\left(\frac{1\text{ cal}}{\text{4}\text{.184 J}}\right)\\ =8.9\times {10}^5\text{ cal}\end{array}$

Determine the temperature change of seawater.

Recall the formula for heat:

$Q=mc\Delta T$

Substitute $0.92\frac{\text{cal}}{\text{gC}°}$ for $c$, $8.9\times {10}^5\text{ cal}$ for $Q$ and $2.06\times {10}^5\text{ g}$ for $m$.

$\begin{array}{c}\left(8.9\times {10}^5\text{ cal}\right)=\left(2.06\times {10}^5\text{ g}\right)\left(0.92\frac{\text{cal}}{\text{gC}°}\right)\Delta T\\ \Delta T=\frac{\left(8.9\times {10}^5\text{ cal}\right)}{\left(2.06\times {10}^5\text{ g}\right)\left(0.92\frac{\text{cal}}{\text{gC}°}\right)}\\ =4.69°\text{C}\end{array}$

Conclusion:

The change in temperature is $4.69°\text{C}$.