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What mass of butter, which has a usable energy content of \(6.0 \mathrm{Cal} / \mathrm{g}(=6000 \mathrm{cal} / \mathrm{g}),\) would be equivalent to the change in gravitational potential energy of a \(73.0 \mathrm{~kg}\) man who ascends from sea level to the top of Mt. Everest, at elevation \(8.84 \mathrm{~km}\) ? Assume that the average \(g\) for the ascent is \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).

Short Answer

Expert verified
Calculate the GPE, convert to calories, and divide by 6000 cal/g to find the mass of butter needed.

Step by step solution

01

Identify the Problem Components

To solve this, we need to calculate the gravitational potential energy (GPE) change and then find the equivalent energy in mass of butter. The given values are the man's mass: 73 kg, height of Mt. Everest: 8840 m, and energy content of butter: 6000 cal/g. Average gravitational acceleration is also provided: 9.80 m/s².
02

Calculate Gravitational Potential Energy

The formula for gravitational potential energy is \( GPE = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height. Substitute into the formula:\[ GPE = 73 \times 9.80 \times 8840 \]Calculating this gives the GPE change.
03

Convert Gravitational Potential Energy into Calories

First, calculate the GPE in joules from Step 2. Next, recall that 1 calorie = 4.184 joules. Convert the gravitational potential energy from joules to calories by dividing the GPE by 4.184.
04

Determine Mass of Butter Needed

Now that the GPE in calories is known, use the energy content of butter to find the required mass. Since butter has 6000 cal/g, divide the total caloric energy (from Step 3) by 6000 cal/g to find the mass of butter.

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Key Concepts

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

Energy Content
Understanding energy content is vital when you need to compare different energy sources. In this exercise, we are dealing with butter's energy content, which is given as 6000 calories per gram. This tells us how much energy we can get from 1 gram of butter.

Energy content is a measure of the potential energy that can be released from a material when it is used or consumed. It's important in designing diets, evaluating fuel sources, and sometimes, like in this exercise, in scientific problems involving energy equivalences. Butter, being rich in fats, has a high energy content, which means it can release a lot of energy when metabolized.
  • Caloric values of foods represent their energy content.
  • Understanding energy content helps us convert energy from various sources easily.
Mass Conversion
Mass conversion involves changing the measure of an amount from one unit to another, often from energy to mass or vice versa. In the context of this problem, it's needed to equate the gravitational potential energy increase to an equivalent mass of butter.

To do this, we first determine the energy in a known form, such as gravitational potential energy in joules, and then convert this into another form, here in calories, using a conversion factor.
  • Using conversion factors is essential: 1 calorie equals 4.184 joules.
  • Once energy in calories is calculated, determine the mass of the source needed to match this energy.
Mass conversion ensures that the energies calculated in different units, such as joules and calories, can be compared or equated to physical amounts like the butter's mass.
Caloric Calculations
Caloric calculations are crucial when converting energy from one form to another. After figuring out the gravitational potential energy (GPE), it needed to be converted into calories for this problem to be relevant to the energy content of butter.

The conversion from joules to calories was made possible by the factor 1 calorie = 4.184 joules. Calculating how many calories are equivalent to the GPE gives a numerical value to compare with butter's energy content.
  • Calculating GPE: Use the formula \( GPE = mgh \) to find energy in Joules.
  • Convert the energy from joules to calories to facilitate comparison with food energy content.
These calculations allow us to physically comprehend how much butter would equate to the energy needed for such a significant altitude change. Understanding these conversions helps demonstrate the interplay between physical energy requirements and food energy values.

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Most popular questions from this chapter

A \(1700 \mathrm{~kg}\) Buick moving at \(83 \mathrm{~km} / \mathrm{h}\) brakes to a stop, at uniform deceleration and without skidding, over a distance of \(93 \mathrm{~m}\). At what average rate is mechanical energy transferred to thermal energy in the brake system?

A cube of edge length \(6.0 \times 10^{-6} \mathrm{~m}\) emissivity \(0.75,\) and temperature \(-100^{\circ} \mathrm{C}\) floats in an environment at \(-150^{\circ} \mathrm{C}\). What is the cube's net thermal radiation transfer rate?

Suppose that you intercept \(5.0 \times 10^{-3}\) of the energy radiated by a hot sphere that has a radius of \(0.020 \mathrm{~m},\) an emissivity of 0.80 , and a surface temperature of \(500 \mathrm{~K}\). How much energy do you intercept in \(2.0 \mathrm{~min} ?\)

Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis (Fig. 18-55). Because the water-ice interface must have a temperature of \(0^{\circ} \mathrm{C}\), the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions. It can lose energy and freeze only by sending energy up (through distance \(L\) ) to the top of the icicle, where the temperature \(T_{r}\) can be below \(0^{\circ} \mathrm{C}\). Take \(L=0.12 \mathrm{~m}\) and \(T_{r}=-5^{\circ} \mathrm{C}\). Assume that the central tube and the upward conduction path both have cross-sectional area \(A .\) In terms of \(A,\) what rate is (a) energy conducted upward and (b) mass converted from liquid to ice at the top of the central tube? (c) At what rate does the top of the tube move downward because of water freezing there? The thermal conductivity of ice is \(0.400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K},\) and the density of liquid water is \(1000 \mathrm{~kg} / \mathrm{m}^{3}\).

Ice has formed on a shallow pond, and a steady state has been reached, with the air above the ice at \(-5.0^{\circ} \mathrm{C}\) and the bottom of the pond at \(4.0^{\circ} \mathrm{C}\). If the total depth of \(i c e+\) water is \(1.4 \mathrm{~m}\), how thick is the ice? (Assume that the thermal conductivities of ice and water are 0.40 and \(0.12 \mathrm{cal} / \mathrm{m} \cdot \mathrm{C}^{\circ} \cdot \mathrm{s},\) respectively. \()\)

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