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Chapter 19: Problem 26

Consider that 200 J of work is done on a system and \(70.0\) cal of thermal energy is trans-ferred out of the system. In the sense of the first law of thermodynamics, what are the values (including algebraic signs) of (a) \(W,(\mathrm{~b})\) \(Q\), and \((\mathrm{c}) \Delta E^{\mathrm{int}} ?\)

Short Answer

Expert verified
W = 200 J, Q = -292.88 J, \( \Delta E_{\text{int}} \) = -92.88 J

Step by step solution

01

- Find the Work Done (W)

The problem states that 200 J of work is done on the system. In thermodynamics, work done _on_ the system is considered positive. Thus, \( W = +200 \text{ J} \)
02

- Determine the Thermal Energy Transferred Out (Q)

The thermal energy transferred out of the system is 70.0 cal. Energy transferred _out_ of the system is negative. Convert calories to joules: 1 cal = 4.184 J. \( Q = -70.0 \text{ cal} \times 4.184 \text{ J/cal} = -292.88 \text{ J} \)
03

- Apply the First Law of Thermodynamics

The first law of thermodynamics is given by the equation \( \Delta E_{\text{int}} = Q + W \). Substituting the values obtained: \( \Delta E_{\text{int}} = -292.88 \text{ J} + 200 \text{ J} \)
04

- Calculate the Change in Internal Energy (\( \Delta E_{\text{int}} \))

Add the values from Step 3 to determine \( \Delta E_{\text{int}} \): \( \Delta E_{\text{int}} = -292.88 \text{ J} + 200 \text{ J} = -92.88 \text{ J} \)

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Key Concepts

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

work done on system
Work done on a system in thermodynamics has a particular sign convention. When work is performed on the system, it means energy is being put into the system, and we consider it as positive. In our exercise, 200 joules of work is done on the system, hence, we denote it as \( W = +200 \, \text{J} \) This work influences the system's internal energy, part of the total energy inside the system, by potentially increasing it if no energy is lost. By using this understanding, identifying how work affects the system and connecting it to internal energy becomes clearer.
thermal energy transfer
Thermal energy transfer, also known as heat transfer, is another crucial component in the first law of thermodynamics. Heat ( Q ) can either enter or leave a system. To keep things specific, thermodynamic conventions state that heat added to the system is positive, while heat taken out (or lost) by the system is negative. In this exercise, 70.0 calories are transferred out of the system. First, we convert calories to joules using the conversion \( 1 \, \text{cal} = 4.184 \, \text{J} \) Hence, the heat transfer is calculated as \( Q = -70.0 \, \text{cal} \times 4.184 \, \text{J/cal} = -292.88 \, \text{J} \) The negative sign indicates the system loses energy. This loss needs to be accounted for when determining the system's internal energy change.
change in internal energy
The change in internal energy ( \( \Delta E_{\text{int}} \) ) of a system represents how the system's total energy alters due to work and heat transfer. According to the first law of thermodynamics, the change in internal energy is given by \( \Delta E_{\text{int}} = Q + W \). In the given exercise, we've already clarified the work done ( W ) is \( +200 \, \text{J} \) and the heat transfer ( Q ) is \( -292.88 \, \text{J} \). Plugging these values into the formula, we get \( \Delta E_{\text{int}} = -292.88 \, \text{J} + 200 \, \text{J} \), which calculates to \( \Delta E_{\text{int}} = -92.88 \, \text{J} \). The negative sign indicates there is an overall decrease in the internal energy of the system. So, while some energy was added to the system via work, more energy was lost due to heat transfer, resulting in a net energy loss.

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Most popular questions from this chapter

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.)

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 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 \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).

A room is lighted by four \(100 \mathrm{~W}\) incandescent lightbulbs. (The power of \(100 \mathrm{~W}\) is the rate at which a bulb converts electrical energy to thermal energy and the energy of visible light.) Assuming that \(90 \%\) of the energy is converted to thermal energy, how much thermal energy is transferred to the room in \(1.00 \mathrm{~h}\) ?

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.
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