Chapter 15, Problem 9PEB

Assume an astronaut at a space station on Mars places a 1 cm thick aluminum griddle with a surface area of 5.00 $\times$ 10² cm² outside to heat it to cook pancakes. Assume the griddle absorbs all of the solar energy, and its initial temperature is 20°C. How long will it take the energy from the Sun to raise the griddle temperature to the desired cooking temperature of 150°C?

To determine

The time taken by energy from the Sun to raise the griddle temperature to the desired cooking temperature of $150°\text{C}$.

Answer to Problem 9PEB

Solution:

$89.76\text{ s}$

Explanation of Solution

Given data:

An astronaut at a space station on Mars places 1 cm thick aluminum griddle with a surface area of $5\times {10}^2{\text{ cm}}^2$ outside to the heat.

Initial temperature is $20°\text{C}$.

Final tempeartute is $150°\text{C}$.

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.

The total energy received is given by the multiplication of the solar energy received, area and the duration of the time.

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

Explanation:

According to the question volume of aluminium is as follows:

$\begin{array}{c}{V}_{\text{aluminum}}=\text{Surface Area }\times \text{ thickness}\\ =\left(5\times {10}^2{\text{ cm}}^2\right)\left(1\text{ cm}\right)\\ =5\times {10}^2{\text{ cm}}^3\end{array}$

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

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

Substitute $2.70\frac{\text{g}}{{\text{cm}}^3}$ for ${\rho }_{\text{aluminum}}$ and $5\times {10}^2{\text{ cm}}^3$ for $V_{\text{aluminum}}$.

$\begin{array}{c}2.70\frac{\text{g}}{{\text{cm}}^3}=\frac{m_{\text{aluminum}}}{5\times {10}^2{\text{ cm}}^3}\\ m_{\text{aluminum}}=\left(2.70\frac{\text{g}}{{\text{cm}}^3}\right)\left(5\times {10}^2{\text{ cm}}^3\right)\\ m_{\text{aluminum}}=1.35\times {10}^3\text{ g}\end{array}$

Determine energy required to heat the griddle to cooking temperature:

Mass of aluminium is $1.35\times {10}^3\text{ g}$.

Specific heat of aluminium is $0.22\frac{\text{cal}}{\text{g}°\text{C}}$.

Initial temperature is $20°\text{C}$.

Final temperature is $150°\text{C}$.

Recall the formula for heat:

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

Substitute $1.35\times {10}^3\text{ g}$ for $m$, $0.22\frac{\text{cal}}{\text{g}°\text{C}}$ for $c$, $20°\text{C}$ for $T_i$ and $150°\text{C}$ for $T_f$.

$\begin{array}{c}Q=\left(1.35\times {10}^3\text{ g}\right)\left(0.22\frac{\text{cal}}{\text{g}°\text{C}}\right)\left(150°\text{C}-20°\text{C}\right)\\ =297\left(130\right)\text{ cal}\\ =3.86\times {10}^4\text{ cal}\end{array}$

Refer the table 15.1 to the value of solar energy received by Mars.

$0.86\frac{\text{cal}}{{\text{cm}}^2\cdot \text{s}}$

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

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

Substitute $0.86\frac{\text{cal}}{{\text{cm}}^2\cdot \text{s}}$ for solar energy received, $5\times {10}^2{\text{ cm}}^2$ for $A$ and $3.86\times {10}^4\text{ cal}$ for $Q$.

$\begin{array}{c}3.86\times {10}^4\text{ cal}=\left(0.86\frac{\text{cal}}{{\text{cm}}^2\cdot \text{s}}\right)\left(5\times {10}^2{\text{ cm}}^2\right)t\\ t=\frac{3.86\times {10}^4\text{ cal}}{\left(0.86\frac{\text{cal}}{{\text{cm}}^2\cdot \text{s}}\right)\left(5\times {10}^2{\text{ cm}}^2\right)}\\ t=89.76\text{ s}\end{array}$

Conclusion:

The time taken energy from the Sun to raise the griddle temperature is $89.76\text{ s}$.