/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 A birch tree loses \(618 \mathrm... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 42

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?

Short Answer

Expert verified
Answer: The rate of heat lost through transpiration is approximately 23.175 J s^-1 (Joules per second).

Step by step solution

01

Convert mass loss to kg per minute

To use the formula for heat, we need to convert the mass loss from mg to kg. Since there are \(1,000,000 \mathrm{mg}\) in a kilogram, we can convert \(618\) mg as follows: $$618 \, \mathrm{mg} = \frac{618}{1,000,000} \, \mathrm{kg} = 6.18 \times 10^{-4} \, \mathrm{kg}$$ Therefore, the mass loss of water per minute in kg is \(6.18 \times 10^{-4} \, \mathrm{kg}\).
02

Calculate the heat of evaporation

Now we can multiply the mass loss by the latent heat of vaporization of water, which is \(2.25 \times 10^6 \, \mathrm{J\,kg^{-1}}\), to find the heat of evaporation, \(Q\): $$Q = m L_v = (6.18 \times 10^{-4} \, \mathrm{kg}) (2.25 \times 10^6 \, \mathrm{J\,kg^{-1}}) = 1390.5 \, \mathrm{J}$$ Thus, the heat of evaporation is \(1390.5 \, \mathrm{J}\).
03

Calculate the rate of heat loss

Finally, we can divide the heat of evaporation by the time (in minutes) to find the rate of heat loss: $$\mathrm{Rate\, of\, heat\, loss} = \frac{Q}{\mathrm{time}} = \frac{1390.5 \, \mathrm{J}}{1\,\mathrm{min}}$$ As there are \(60\) seconds in a minute, we convert the heat loss rate to Joules per second: $$\mathrm{Rate\, of\, heat\, loss} = \frac{1390.5 \, \mathrm{J}}{60\,\mathrm{s}} = 23.175 \, \mathrm{J\,s^{-1}}$$ Hence, the rate of heat lost through transpiration is approximately \(23.175 \, \mathrm{J\,s^{-1}}\) (Joules per second).

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Most popular questions from this chapter

A chamber with a fixed volume of \(1.0 \mathrm{m}^{3}\) contains a monatomic gas at \(3.00 \times 10^{2} \mathrm{K} .\) The chamber is heated to a temperature of \(4.00 \times 10^{2} \mathrm{K}\). This operation requires \(10.0 \mathrm{J}\) of heat. (Assume all the energy is transferred to the gas.) How many gas molecules are in the chamber?
A \(0.400-\mathrm{kg}\) aluminum teakettle contains \(2.00 \mathrm{kg}\) of water at \(15.0^{\circ} \mathrm{C} .\) How much heat is required to raise the temperature of the water (and kettle) to \(100.0^{\circ} \mathrm{C} ?\)
For a cheetah, \(70.0 \%\) of the energy expended during exertion is internal work done on the cheetah's system and is dissipated within his body; for a dog only \(5.00 \%\) of the energy expended is dissipated within the dog's body. Assume that both animals expend the same total amount of energy during exertion, both have the same heat capacity, and the cheetah is 2.00 times as heavy as the dog. (a) How much higher is the temperature change of the cheetah compared to the temperature change of the dog? (b) If they both start out at an initial temperature of \(35.0^{\circ} \mathrm{C},\) and the cheetah has a temperature of \(40.0^{\circ} \mathrm{C}\) after the exertion, what is the final temperature of the dog? Which animal probably has more endurance? Explain.

A student wants to lose some weight. He knows that rigorous aerobic activity uses about \(700 \mathrm{kcal} / \mathrm{h}(2900 \mathrm{kJ} / \mathrm{h})\) and that it takes about 2000 kcal per day \((8400 \mathrm{kJ})\) just to support necessary biological functions, including keeping the body warm. He decides to burn calories faster simply by sitting naked in a \(16^{\circ} \mathrm{C}\) room and letting his body radiate calories away. His body has a surface area of about \(1.7 \mathrm{m}^{2}\) and his skin temperature is \(35^{\circ} \mathrm{C} .\) Assuming an emissivity of \(1.0,\) at what rate (in \(\mathrm{kcal} / \mathrm{h}\) ) will this student "burn" calories?
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