A leaf of area \({\bf{40}}\;{\bf{c}}{{\bf{m}}^{\bf{2}}}\) and mass \({\bf{4}}{\bf{.5 \times 1}}{{\bf{0}}^{{\bf{ - 4}}}}\;{\bf{kg}}\) directly faces the Sun on a clear day. The leaf has an emissivity of 0.85 and a specific heat of \({\bf{0}}{\bf{.80}}\;{\bf{kcal/kg}} \cdot {\bf{K}}\) (a) Estimate the energy absorbed per second by the leaf from the Sun, and then (b) estimate the rate of rise of the leaf’s temperature. (c) Will the temperature rise continue for hours? Why or why not? (d) Calculate the temperature the leaf would reach if it lost all its heat by radiation to the surroundings at 24°C. (e) In what other ways can the heat be dissipated by the leaf?

The Earth’s surface receives energy from the Sun at a rate of about\({\bf{1000}}\;{\bf{W/}}{{\bf{m}}^{\bf{2}}}\)on a clear day. However, an object of emissivity \(\varepsilon \),with area A,and facing towards the Sun absorbs energy from the Sun at a rate of about

\(\frac{Q}{t} = \left( {1000\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}} \right)\varepsilon A\cos \theta \). … (i)

Here, \(A\cos \theta \) is the effective area at right angles to the Sun if the rays make an angle \(\theta \) with a line perpendicular to the area.

Mass of the leaf is \(m = 4.5 \times {10^{ - 4}}\;{\rm{kg}}\).

Area of the leaf is \(A = 40\;{\rm{c}}{{\rm{m}}^{\rm{2}}} = 40 \times {10^{ - 4}}\;{{\rm{m}}^2}\).

Emissivity of the leaf is \(\varepsilon = 0.85\).

Temperature of the surroundings is \({T_2} = 24^\circ {\rm{C}}\).

The specific heat of the leaf is \(\begin{aligned}{c}c = \left( {0.80\;{\rm{kcal/kg}} \cdot {\rm{K}}} \right) \times \left( {\frac{{4186\;{\rm{J}}}}{{1\;{\rm{kcal}}}}} \right)\\ = 3348.8\;{\rm{J/kg}} \cdot {\rm{K}}{\rm{.}}\end{aligned}\)

Since the leaf faces the Sun directly, the angle \(\theta \) is zero.

From equation (i), the energy absorbed by the leaf from the Sun is

\(\begin{aligned}{c}\frac{Q}{t} = \left( {1000\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}} \right)\varepsilon A\cos 0^\circ \\ = \left( {1000\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}} \right)\left( {0.85} \right)\left( {40 \times {{10}^{ - 4}}\;{{\rm{m}}^2}} \right) \times 1\\ = 3.4\;{\rm{W}}{\rm{.}}\end{aligned}\)

Thus, the energy absorbed per second by the leaf from the Sun is 3.4 W.

Suppose all of the energy absorbed by the leaf from the Sun (Q) is used in raising its temperature by \(\Delta T\). Then the amount of heat required to raise the temperature of the leaf by\(\Delta T\)must be equal to the energy absorbed from the Sun.

The amount of heat required to raise the temperature is

\(Q = mc\Delta T\).

So, the rise in temperature of the leaf is

\(\Delta T = \frac{Q}{{mc}}\).

Divide both sides by time t to get the rate of rise of the leaf’s temperature.

\(\begin{aligned}{c}\frac{{\Delta T}}{t} = \frac{1}{{mc}} \times \frac{Q}{t}\\ = \frac{{3.4\;W}}{{\left( {4.5 \times {{10}^{ - 4}}\;{\rm{kg}}} \right)\left( {3348.8\;{\rm{J/kg}} \cdot {\rm{K}}} \right)}}\\ = 2.256\;{\rm{K/s}}\\ \approx 2.3\;{\rm{K/s}}\;{\rm{or}}\;2.3^\circ {\rm{C/s}}\end{aligned}\)

Thus, the rate of rise of the leaf’s temperature is about \(2.3^\circ {\rm{C/s}}\).

The temperature of the leaf rises by about \(2.3^\circ {\rm{C}}\) in 1 second. Thus, the rise in temperature of the leaf in 1 hour is

\(\begin{aligned}{c}\Delta T = \;2.3^\circ {\rm{C/s}} \times t\\\Delta {\rm{T}} = \left( {2.3^\circ {\rm{C/s}}} \right) \times \left( {\frac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}}} \right)\\ = 8280^\circ {\rm{C}}{\rm{.}}\end{aligned}\)

The temperature of the leaf would rise by more than \(8000^\circ {\rm{C}}\) in just one hour. This value of temperature is too high and it will burn the leaf before the end of the hour. Therefore, the rise in temperature will not continue for hours.

Since both the surfaces of the leaf lose heat by radiation to the surroundings, the area of the leaf radiating heat is

\(A' = 2A\).

If temperature of the leaf is \({T_1}\), then the rate of loss of heat by the leaf is

\({\left( {\frac{Q}{t}} \right)^\prime } = \varepsilon \sigma A'\left( {T_1^4 - T_2^4} \right)\). … (ii)

Here, \(\sigma \) is the Stefan-Boltzmann constant with the value \(5.67 \times {10^{ - 8}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}} \cdot {{\rm{K}}^4}\).

The rate of loss of heat must be equal to the rate of absorption of heat.

Therefore,

\(\begin{aligned}{c}{\left( {\frac{Q}{t}} \right)^\prime } = \frac{Q}{t}\\\varepsilon \sigma A'\left( {T_1^4 - T_2^4} \right) = 3.4\;{\rm{W}}\\\left( {T_1^4 - T_2^4} \right) = \frac{{3.4\;{\rm{W}}}}{{\varepsilon \sigma A'}}\\{T_1} = {\left( {\frac{{3.4\;{\rm{W}}}}{{\varepsilon \sigma A'}} + T_2^4} \right)^{\frac{1}{4}}}.\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}{T_1} = {\left( {\frac{{3.4\;{\rm{W}}}}{{\left( {{\rm{0}}{\rm{.85}}} \right)\left( {5.67 \times {{10}^{ - 8}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}} \cdot {{\rm{K}}^4}} \right)\left( {2 \times 40 \times {{10}^{ - 4}}\;{{\rm{m}}^2}} \right)}} + {{\left( {297\;{\rm{K}}} \right)}^4}} \right)^{\frac{1}{4}}}\\ = 359\;{\rm{K}}\\ = \left( {{\rm{359 - 273}}} \right)^\circ {\rm{C}}\\ = 8{\rm{6}}^\circ {\rm{C}}\end{aligned}\)

Thus, the leaf would reach a temperature of \(8{\rm{6}}^\circ {\rm{C}}\).

The leaf can also dissipate heat by evaporation, conduction, and convection.

Evaporation: The leaf can lose its heat by losing water to the surroundings.

Conduction: The leaf can directly lose heat to the air in contact with it. Since the air is at a lower temperature, heat will directly flow from the leaf at a higher temperature to the air at a lower temperature.

Convection: Since the air around the leaf moves continuously, the leaf can lose heat by convection.