Chapter 15, Problem 8PEB

A 1 cm thick piece of lead with a surface area of 160 cm² is placed on the surface of Mercury for 1 hour during the time the planet receives its peak solar energy. Assume the lead absorbs all of the solar energy, and its temperature when placed is at absolute zero. Will the energy received raise the temperature of the lead to its melting point of 327°C? If so, determine the mass of lead that melts. (Assume L_f of lead is 5 cal/g.)

To determine

The mass of lead melts, if the energy received raise the temperature of the lead to its melting point of $327°\text{C}$.

Answer to Problem 8PEB

Solution:

$1.5\times {10}^6\text{ g}$

Explanation of Solution

Given data:

A $1\text{ cm}$ thick piece of lead with a surface area of $160{\text{ cm}}^2$ is placed on the surface of mercury for 1 hour during that time the planet receives its peak solar energy.

The energy received raise temperature of the lead to its melting pon of $327°\text{C}$.

Latent heat of lead is $5\frac{\text{cal}}{\text{g}}$.

Formula used:

Write the expression of density.

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

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

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.

Write the equation for latent heat $\left(L_{\text{f}}\right)$ relationship:

$Q=m{L}_{\text{f}}$

Here, $Q$ is the energy and $m$ is mass.

Explanation:

According to the question volume of lead is as follows:

$\begin{array}{c}{V}_{\text{lead}}=\left(160{\text{ cm}}^2\right)\left(1\text{ cm}\right)\\ =160{\text{ cm}}^3\end{array}$

Recall the expression of density to find the mass of lead.

${\rho }_{\text{lead}}=\frac{m_{\text{lead}}}{V_{\text{lead}}}$

Substitute $11.4\frac{\text{g}}{{\text{cm}}^3}$ for ${\rho }_{\text{lead}}$ and $160{\text{ cm}}^3$ for $V_{\text{lead}}$.

$\begin{array}{c}11.4\frac{\text{g}}{{\text{cm}}^3}=\frac{m_{\text{lead}}}{160{\text{ cm}}^3}\\ m_{\text{lead}}=\left(11.4\frac{\text{g}}{{\text{cm}}^3}\right)\left(160{\text{ cm}}^3\right)\\ m_{\text{lead}}=1.8\times {10}^3\text{ g}\end{array}$

Determine the energy required to heat the lead to its melting point:

Mass of lead is $1.8\times {10}^3\text{ g}$.

Specific heat of mercury is $0.0305\frac{\text{cal}}{\text{g}°\text{C}}$.

Initial temperature is $-273°\text{C}$.

Final temperature is $327°\text{C}$.

Recall the formula for heat:

$Q=mc\left(T_f-T_i\right)$

Substitute $1.8\times {10}^3\text{ g}$ for $m$, $0.0305\frac{\text{cal}}{\text{g}°\text{C}}$ for $c$, $-273°\text{C}$ for $T_i$ and $327°\text{C}$ for $T_f$.

$\begin{array}{c}Q=\left(1.8\times {10}^3\text{ g}\right)\left(0.0305\frac{\text{cal}}{\text{g}°\text{C}}\right)\left(327°\text{C}-\left(-273°\text{C}\right)\right)\\ =54.9\left(600\right)\text{ cal}\\ =3.3\times {10}^4\text{ cal}\end{array}$

Refer the table 15.1 to the value of mercury of solar energy received.

$13.4\frac{\text{cal}}{{\text{cm}}^2\cdot \text{s}}$

Convert time to seconds:

$\begin{array}{c}1\text{ h}=1\text{ h}\left(\frac{3.6\times {10}^3\text{ s}}{1\text{ h}}\right)\\ =3.6\times {10}^3\text{ s}\end{array}$

Determine the energy received, use solar energy received multiply it by the area and the duration of the time.

$Q=\left(\text{Solar energy received}\right)\left(At\right)$

Substitute $13.4\frac{\text{cal}}{{\text{cm}}^2\cdot \text{s}}$ for solar energy received, $160{\text{ cm}}^2$ for $A$ and $3.6\times {10}^3\text{ s}$ for $t$.

$\begin{array}{c}Q=\left(13.4\frac{\text{cal}}{{\text{cm}}^2\cdot \text{s}}\right)\left(160{\text{ cm}}^2\right)\left(3.6\times {10}^3\text{ s}\right)\\ =7.7\times {10}^6\text{ cal}\end{array}$

Compare the energy received to the energy required to heat the lead and its melting point:

$7.7\times {10}^6\text{ cal}>3.3\times {10}^4\text{ cal}$

Therefore, the lead will reach its melting point.

Determine the excess energy:

$\text{Excess energy available}=\text{Energy received}-\text{Energy used}$

Substitute $7.7\times {10}^6\text{ cal}$ for energy received and $3.3\times {10}^4\text{ cal}$ for energy used.

$\begin{array}{c}\text{Excess energy available}=7.7\times {10}^6\text{ cal}-3.3\times {10}^4\text{ cal}\\ =\text{7}\text{.66}\times {\text{10}}^6\text{ cal}\end{array}$

Recall the equation for latent heat $\left(L_{\text{f}}\right)$ relationship:

$Q=m{L}_{\text{f}}$

Substitute $\text{7}\text{.66}\times {\text{10}}^6\text{ cal}$ for $Q$ and $5\frac{\text{cal}}{\text{g}}$ for $L_{\text{f}}$.

$\begin{array}{c}\text{7}\text{.66}\times {\text{10}}^6\text{ cal}=m\left(5\frac{\text{cal}}{\text{g}}\right)\\ m=\frac{\text{7}\text{.66}\times {\text{10}}^6\text{ cal}}{5\frac{\text{cal}}{\text{g}}}\\ m=1.5\times {10}^6\text{ g}\end{array}$

Conclusion:

The mass of lead melts are $1.5\times {10}^6\text{ g}$.