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Chapter 15: 15Q (page 412)


Question:The first law of thermodynamics is sometimes whimsically stated as, 鈥淵ou can鈥檛 get something for nothing,鈥� and the second law as, 鈥淵ou can鈥檛 even break even.鈥� Explain how these statements could be equivalent to the formal statements.

Short Answer

Expert verified

The statement 鈥淵ou can鈥檛 get something for nothing鈥� implies that energy is conserved. The other statement, 鈥淵ou can鈥檛 even break even,鈥� tells us that it is impossible to have a 100% effective heat engine.

Step by step solution

01

Understanding the second law of thermodynamics 

From the second law of thermodynamics, it can be obtained that there is no such heat engine that gives 100 percent efficiency without any losses or limitations.

02

Explanation of the statement “You can’t get something for nothing” in terms of the first law of thermodynamics

The statement 鈥淵ou can鈥檛 get something for nothing鈥� correlates to the first law of thermodynamics because it could be a way of saying that energy is conserved.

The work performed by a specific system does require heat from an outside source or its internal energy. So, to obtain work done, something is applied by the system.

Thus, it is equivalent to the formal statement of the first law of thermodynamics.

03

Explanation of the statement “You can’t even break even” in terms of the second law of thermodynamics

The statement, 鈥淵ou can鈥檛 even break even,鈥� relates to the second law of thermodynamics because it indicates that a certain heat engine cannot produce 100 percent efficiency.

For example, if the engine takes heat through the input, it is impractical to give an exact amount of energy at the output because some energy will be lost.

Thus, it is equivalent to the formal statement of the second law of thermodynamics.

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Most popular questions from this chapter

Question:(II) A four-cylinder gasoline engine has an efficiency of 0.22 and delivers 180 J of work per cycle per cylinder. If the engine runs at 25 cycles per second (1500 rpm), determine (a) the work done per second, and (b) the total heat input per second from the gasoline. (c) If the energy content of gasoline is 130 MJ per gallon, how long does one gallon last?

Calculate the work done by an ideal gas while going from state A to state C in Fig. 15鈥�28 for each of the following processes:

(a) ADC,

(b) ABC, and

(c) AC directly.

FIGURE 15鈥�28

Problem 68

(II) When\({\bf{5}}{\bf{.80 \times 1}}{{\bf{0}}{\bf{5}}}\;{\bf{J}}\)of heat is added to a gas enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure, the volume is observed to increase from\({\bf{1}}{\bf{.9}}\;{{\bf{m}}{\bf{3}}}\)to\({\bf{4}}{\bf{.1}}\;{{\bf{m}}{\bf{3}}}\). Calculate

(a) the work done by the gas, and

(b) the change in internal energy of the gas.

(c) Graph this process on a PV diagram.

Question: (a) At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 750掳C and 440掳C, and of the second 415掳C and 270掳C. If the heat of combustion of coal is \({\bf{2}}{\bf{.8 \times 1}}{{\bf{0}}^{\bf{7}}}\;{{\bf{J}} \mathord{\left/{\vphantom {{\bf{J}} {{\bf{kg}}}}} \right.} {{\bf{kg}}}}\) at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65% of the ideal (Carnot) efficiency. (b) Water is used to cool the power plant. If the water temperature is allowed to increase by no more than 4.5 C掳, estimate how much water must pass through the plant per hour.

Question: (II) An ideal gas expands at a constant total pressure of 3.0 atm from 410 mL to 690 mL. Heat then flows out of the gas at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value. Calculate (a) the total work done by the gas in the process, and (b) the total heat flow into the gas.
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