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Chapter 19: Problem 9

An energetic athlete can use up all the energy from a diet of 4000 food calories/day where a food calorie = 1000 cal. If he were to use up this energy at a steady rate, how would his rate of energy use compare with the power of a \(100 \mathrm{~W}\) bulb? (The power of \(100 \mathrm{~W}\) is the rate at which the bulb converts electrical energy to thermal energy and the energy of visible light.)

Short Answer

Expert verified
The athlete's rate of energy use is 193.75 W, which is greater than the power of a 100 W bulb.

Step by step solution

01

- Convert Food Calories to Joules

First, determine the total energy in joules. Given that 1 food calorie is equivalent to 1000 regular calories and 1 calorie equals 4.184 joules, convert the daily energy consumption into joules. \[ 4000 \text{ food calories/day} = 4000 \times 1000 \text{ cal/day} \] \[ 4000000 \text{ cal/day} \times 4.184 \text{ J/cal} = 16736000 \text{ J/day} \]
02

- Convert Daily Energy to Energy per Second

Since power is the rate of energy use per unit of time in seconds, convert the daily energy usage to joules per second. There are 86400 seconds in a day. \[ \frac{16736000 \text{ J/day}}{86400 \text{ s/day}} \ \frac{16736000}{86400} \text{ J/s} = 193.75 \text{ W} \]
03

- Compare with the Power of a 100 W Bulb

Now, compare the rate of energy use (power) to the power of a 100 W bulb. \[ 193.75 \text{ W} \text{ (athlete's power)} > 100 \text{ W} \text{ (bulb's power)} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calories to Joules Conversion
Understanding how to convert calories to joules is essential. Energy in food is usually measured in food calories, but scientific calculations often require joules. Remember, 1 food calorie equals 1000 regular calories. To convert these calories into joules, we use the conversion factor where 1 regular calorie equals 4.184 joules. For instance, with an intake of 4000 food calories per day, this translates to 4000 x 1000 calories. If you multiply this by 4.184, you get the total energy in joules. So, 4000 food calories per day converts into:
\[4000 \ \text{food calories/day} \times 1000 \ \text{calories/food calorie} \times 4.184 \ \text{joules/calorie} = 16736000 \ \text{joules/day} \]
This equation helps in understanding how much energy an athlete consumes and is crucial for further calculations.
Power Calculation
Power refers to the rate at which energy is used or transferred. It's essential to convert the daily energy intake into energy per second because power is measured in watts, equivalent to joules per second. To find this, you need to know the total number of seconds in a day, which is 86400. Dividing the total energy intake in joules by the total seconds in a day gives you the average power the athlete uses:
\[\frac{16736000 \ \text{joules/day}}{86400 \ \text{seconds/day}} = 193.75 \ \text{watts} \]
So, the athlete uses energy at a rate of 193.75 watts. This step is key to converting long-term energy consumption into an immediate power rate, making it easier to compare with other power usages.
Energy Rate Comparison
Comparing energy usage rates involves understanding different power outputs. In this case, an average athlete's energy usage is compared with a 100-watt light bulb. From the calculations, the athlete鈥檚 power usage is found to be 193.75 watts. When compared to the light bulb's 100-watt power, it becomes clear:
\[193.75 \ \text{watts} \ (\text{athlete's power}) > 100 \ \text{watts} \ (\text{bulb's power}) \]
This means the athlete uses energy at a much higher rate than the light bulb. Understanding such comparisons can help grasp how much energy is consumed by various activities or devices, emphasizing the energy demands of physical exertion as compared to everyday electrical appliances.
It's important to visualize these differences to appreciate the physical demands on an athlete and the efficient conversion of energy in more sedentary activities.

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Most popular questions from this chapter

In a solar water heater, radiant energy from the Sun is transferred to water that circulates through tubes in a rooftop collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes; this water is pumped into a holding tank. Assume that the efficiency of the overall system is \(20 \%\) (that is, \(80 \%\) of the incident solar energy is lost from the system). What collector area is necessary to raise the temperature of \(200 \mathrm{~L}\) of water in the tank from \(20^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) in \(1.0 \mathrm{~h}\) when the intensity of incident sunlight is \(700 \mathrm{~W} / \mathrm{m}^{2} ?\)

A tank of water has been outdoors in cold weather, and a slab of ice \(5.0 \mathrm{~cm}\) thick has formed on its surface (Fig. 19-42). The air above the ice is at \(-10^{\circ} \mathrm{C}\). Calculate the rate of formation of ice (in centimeters per hour) on the ice slab. Take the thermal conductivity and density of ice to be \(0.0040 \mathrm{cal} / \mathrm{s} \cdot \mathrm{cm} \cdot \mathrm{C}^{\circ}\) and \(0.92 \mathrm{~g} / \mathrm{cm}^{3}\). Assume that energy is not transferred through the walls or bottom of the tank.

300 F Club You can join the semi-secret "300 F" club at the Amundsen-Scott South Pole Station only when the outside temperature is below \(-70^{\circ} \mathrm{C}\). On such a day, you first bask in a hot sauna and then run outside wearing only your shoes. (This is, of course, extremely dangerous, but the rite is effectively a protest against the constant danger of the winter cold at the south pole.) Assume that when you step out of the sauna, your skin temperature is \(102^{\circ} \mathrm{F}\) and the walls, ceiling, and floor of the sauna room have a temperature of \(30^{\circ} \mathrm{C}\). Estimate your surface area, and take your skin emissivity to be \(0.80 .\) (a) What is the approximate net rate \(P^{\text {net }}\) at which you lose energy via thermal radiation transfer to the room? Next, assume that when you are outside half your surface area transfers thermal radiation to the sky at a temperature of \(-25^{\circ} \mathrm{C}\) and the other half transfers thermal radiation to the snow and ground at a temperature of \(-80^{\circ} \mathrm{C}\). What is the approximate net rate at which you lose energy via thermal radiation exchanges with (b) the sky and (c) the snow and ground?

How much water remains unfrozen after \(50.2 \mathrm{~kJ}\) of thermal energy is transferred from \(260 \mathrm{~g}\) of liquid water initially at its freezing point?

A sphere of radius \(0.500 \mathrm{~m}\), temperature \(27.0^{\circ} \mathrm{C}\), and emissivity \(0.850\) is located in an environment of temperature \(77.0^{\circ} \mathrm{C}\). At what rate does the sphere (a) emit and (b) absorb thermal radiation? (c) What is the sphere's net rate of energy exchange?
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