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Chapter 19: Problem 12

How many grams 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 value of \(g\) is \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).

Short Answer

Expert verified
252.5 grams of butter.

Step by step solution

01

- Calculate the change in gravitational potential energy

First, use the formula to calculate the change in gravitational potential energy: \[ \Delta U = mgh \] where: \( m \) is the mass of the person, \( 73.0 \) kg, \( g \) is the acceleration due to gravity, \( 9.80 \) m/s\textsuperscript{2}, and \( h \) is the height, \( 8.84 \) km (converted to meters). Now, convert the height from kilometers to meters: \[ 8.84 \text{ km} = 8.84 \times 10^3 = 8840 \text{ m} \]Next, calculate the change in potential energy: \[ \Delta U = 73.0 \times 9.80 \times 8840 \] \[ \Delta U = 6338352 \text{ J} \approx 6.34 \times 10^6 \text{ J} \]
02

- Convert the energy from Joules to calories

Use the conversion factor: \[ 1 \text{ cal} = 4.184 \text{ J} \] Converting the gravitational potential energy to calories: \[ 6.34 \times 10^6 \text{ J} \div 4.184 \text{ J/cal} = 1.515 脳 10^6 \text{ cal} \]
03

- Calculate the grams of butter needed to provide the same energy

Given the energy content of butter is \( 6000 \) cal/g, find the mass in grams: \[ \text{Mass of butter} = \frac{1.515 \times 10^6 \text{ cal}}{6000 \text{ cal/g}} \] \[ \text{Mass of butter} \approx 252.5 \text{ grams} \]

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Key Concepts

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

energy conversion
Energy conversion is a crucial concept in physics. It involves changing one form of energy into another. In our exercise, the man's gravitational potential energy is converted into electrical or mechanical energy in a lab setting. When he climbs Mt. Everest, he gains gravitational potential energy because of his elevation. This energy can be turned into other forms meaningfully. For example, in power plants, mechanical energy is converted into electrical energy.

Understanding energy conversion helps engineers design efficient energy systems, keeping sustainability in mind. Our bodies also convert the chemical energy from food into the mechanical energy needed for movement. It's like how the butter we calculated converts chemical energy from calories into usable energy for our body.
calories to joules
Converting calories to joules is essential in physics problems, especially those involving nutrition and energy. Calories (Cal) measure the energy content in food, while joules (J) are the SI unit for energy. For instance, the energy content in butter can be calculated for physics equations. You can convert calories to joules using the conversion factor:
  • 1 calorie = 4.184 joules.

To make it easier, if you know how many calories a food item has, you can multiply that number by 4.184 to get the energy in joules.
In our exercise, we converted the energy from joules to calories. This helped us find out how much butter the man needs to climb Mt. Everest. This conversion is essential in various fields, like diet planning, where understanding food energy in joules can help balance physical activity and energy intake.
physics problem solving
Solving physics problems involves breaking down the problem into manageable steps. Here, we used the gravitational potential energy formula:

\(\Delta U = mgh \)

where:
  • m is the mass,
  • g is gravity,
  • h is height.
Each component is essential for accurate calculations.

Next, we applied unit conversion from kilometers to meters to get the height in standard units. Afterward, we calculated the potential energy change by multiplying mass, gravity, and height. Finally, converting the energy into calories helped use the food energy content to find the grams of butter. This step-by-step approach is key to solving any complex physics problem accurately.
mass-energy equivalence
Mass-energy equivalence, formulated by Einstein's equation, \(E=mc^2\), shows the relationship between mass and energy. In our context, it isn't directly used, but it underscores the principle that mass can be thought of as a form of energy.

In climbing Mt. Everest, the man's mass gains gravitational potential energy due to elevation. Similarly, high-energy food like butter provides calories that can be converted into useful energy. This principle helps explain how energy is stored and transferred, both in physical systems and biological ones. Understanding mass-energy equivalence is fundamental in modern physics, influencing everything from nuclear energy to cosmology.

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Most popular questions from this chapter

Show that when the temperature of a liquid in a barometer changes by \(\Delta T\) and the pressure is constant, the liquid's height \(h\) changes by \(\Delta h=\beta h \Delta T\), where \(\beta\) is the coefficient of volume expansion. Neglect the expansion of the glass tube.

. Without a Spacesuit If you were to walk briefly in space without a spacesuit while far from the Sun (as an astronaut does in the movie 2001 ), you would feel the cold of space-while you radiated thermal energy, you would absorb almost none from your environment. (a) At what rate would you lose thermal energy? (h) How much thermal energy would you lose in \(30 \mathrm{~s}\) ? Assume that your emissivity is \(0.90\), and estimate other data needed in the calculations.

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?

For each of the situations described below, the object considered is undergoing some changes. Among the possible changes you should consider are: (Q) The object is absorbing or giving off thermal energy. (T) The object's temperature is changing. \(\left(E^{\text {int }}\right)\) The object's internal energy is changing. (W) The object is doing mechanical work or having work done on it. For each of the situations described below, identify which of the four changes are taking place and write as many of the letters \(\mathrm{Q}\), \(\mathrm{T}, E^{\text {int }}, \mathrm{W}\), (or none) as are appropriate. (a) A cylinder with a piston on top contains a compressed gas and is sitting on a thermal reservoir (a large iron block). After everything has come to thermal equilibrium, the piston is moved upward somewhat (very slowly). The object to be considered is the gas in the cylinder. (b) Consider the same cylinder as in part (a), but it is wrapped in styrofoam, a very good thermal insulator, instead of sitting on a thermal reservoir. The piston is pressed downward (again, very slowly), compressing the gas. The object to be considered is the

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?
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