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

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

Step by step solution

01

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 = \((9.00 \times 10^{2} \mathrm{W/m^2}) \times 0.700 = 6.30 \times 10^{2} \mathrm{W/m^2}\)
02

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 = \((6.30 \times 10^{2} \mathrm{W/m^2}) \times (5.00 \times 10^{-3} \mathrm{m^2}) = 3.15 \mathrm{W}\)
03

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 \(1.70 \mathrm{^{\circ}C/s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A piece of gold of mass \(0.250 \mathrm{kg}\) and at a temperature of \(75.0^{\circ} \mathrm{C}\) is placed into a \(1.500-\mathrm{kg}\) copper pot containing \(0.500 \mathrm{L}\) of water. The pot and water are at $22.0^{\circ} \mathrm{C}$ before the gold is added. What is the final temperature of the water?
A stainless steel saucepan, with a base that is made of \(0.350-\mathrm{cm}-\) thick steel \([\kappa=46.0 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})]\) fused to a \(0.150-\mathrm{cm}\) thickness of copper $[\kappa=401 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})],\( sits on a ceramic heating element at \)104.00^{\circ} \mathrm{C} .\( The diameter of the pan is \)18.0 \mathrm{cm}$ and it contains boiling water at \(100.00^{\circ} \mathrm{C} .\) (a) If the copper-clad bottom is touching the heat source, what is the temperature at the copper-steel interface? (b) At what rate will the water evaporate from the pan?
A birch tree loses \(618 \mathrm{mg}\) of water per minute through transpiration (evaporation of water through stomatal pores). What is the rate of heat lost through transpiration?
What is the heat capacity of a system consisting of (a) a \(0.450-\mathrm{kg}\) brass cup filled with \(0.050 \mathrm{kg}\) of water? (b) \(7.5 \mathrm{kg}\) of water in a 0.75 -kg aluminum bucket?
A hiker is wearing wool clothing of \(0.50-\mathrm{cm}\) thickness to keep warm. Her skin temperature is \(35^{\circ} \mathrm{C}\) and the outside temperature is \(4.0^{\circ} \mathrm{C} .\) Her body surface area is \(1.2 \mathrm{m}^{2} .\) (a) If the thermal conductivity of wool is $0.040 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ what is the rate of heat conduction through her clothing? (b) If the hiker is caught in a rainstorm, the thermal conductivity of the soaked wool increases to \(0.60 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) (that of water). Now what is the rate of heat conduction?
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free