Question

If the total power per unit area from the Sun incident on a horizontal leaf is \(9.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2},\) and we assume that \(70.0 \%\) of this energy goes into heating the leaf, what would be the rate of temperature rise of the leaf? The specific heat of the leaf is $3.70 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right),$ the leaf's area is \(5.00 \times 10^{-3} \mathrm{m}^{2},\) and its mass is $0.500 \mathrm{g}$.

Short answer

Answer: The rate of temperature rise of the leaf is approximately \(1.70 \mathrm{^{\circ}C/s}\).

Step-by-step solution

Calculate the total power absorbed by the leaf

Since we know the total power per unit area from the Sun and the percentage of energy that goes into heating the leaf, we can calculate the total power absorbed by the leaf. Power per unit area = \(9.00 \times 10^{2} \mathrm{W/m^2}\) Percentage of energy for heating = \(70.0 \% = 0.700\) Total power absorbed by the leaf = Power per unit area \(\times\) Percentage of energy for heating Total power absorbed by the leaf = \((9.00 \times 10^{2} \mathrm{W/m^2}) \times 0.700 = 6.30 \times 10^{2} \mathrm{W/m^2}\)

Calculate the power absorbed by the leaf for its area

Now we have the total power absorbed by the leaf per unit area, we will multiply it by the area of the leaf to obtain the power absorbed by the leaf. Leaf's area = \(5.00 \times 10^{-3} \mathrm{m^2}\) Power absorbed by the leaf = Total power absorbed by the leaf \(\times\) Leaf's area Power absorbed by the leaf = \((6.30 \times 10^{2} \mathrm{W/m^2}) \times (5.00 \times 10^{-3} \mathrm{m^2}) = 3.15 \mathrm{W}\)

Find the rate of temperature rise

The energy absorbed by the leaf is being used to increase its temperature. To find the rate of temperature rise, we can use the following formula: Rate of temperature rise = \(\frac{\text{Power absorbed by the leaf}}{\text{Specific heat }\times\text{ Mass of the leaf}}\) Specific heat = \(3.70 \mathrm{kJ/kg \cdot ^{\circ}C} = 3.70 \times 10^3 \mathrm{J/kg\cdot^{\circ}C}\) Mass of the leaf = \(0.500 \mathrm{g} = 0.500 \times 10^{-3} \mathrm{kg}\) Rate of temperature rise = \(\frac{3.15 \mathrm{W}}{(3.70 \times 10^3 \mathrm{J/kg\cdot^{\circ}C})\times(0.500\times 10^{-3} \mathrm{kg})} = 1.698 \times 10^{0} \mathrm{^{\circ}C/s}\) The rate of temperature rise of the leaf is approximately \(1.70 \mathrm{^{\circ}C/s}\).