Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 19, Problem 37P
(a)
To determine
The efficiency of
(b)
To determine
The reason for that no warm blooded organisms have evolved heat engine to increase their heat energies.
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Students have asked these similar questions
Second Law of
thermodynamics.
An engine in performing 196 J of work discharges 34 J of heat. What is its
efficiency? (Give your answer as a number with two decimals).
1. Suppose a woman does 600 J of work and -9400 J of heat transfer occurs into the environment in
the process.
(a) What is the decrease in her internal energy, assuming no change in temperature or consumption
of food? (That is, there is no other energy transfer.)
J
AE int
(b) The internal energy is stored energy due to food intake. Treating the change in internal energy as
the input energy and work done as output, what is her efficiency?
Efficiency, Eff:
(c) What physics law did you use in this problem?
O Zeroth Law of Thermodynamics
O First Law of Thermodynamics
O Second Law of Thermodynamics
1. (a) What is the best coefficient of performance for a refrigerator that cools an environment at -26°
C and has heat transfer to another environment at 49 ° C?
COP
ref
(b) How much work must be done for a heat transfer of 4186 kJ from the cold environment?
W =
(c) What is the cost of doing this if the work costs 10.0 cents per 3.6 x 106 J (a kilowatt-hour)?
Cost in cents =
(d) How many kJ of heat transfer, Qh occurs into the warm environment?
Qn
kj
=
Think about what type of refrigerator might operate between these temperatures.
Hint: Use the appropriate formula for a refrigerator which is different from a heat pump.
Chapter 19 Solutions
Physics for Scientists and Engineers
Ch. 19 - Prob. 1PCh. 19 - Prob. 2PCh. 19 - Prob. 3PCh. 19 - Prob. 4PCh. 19 - Prob. 5PCh. 19 - Prob. 6PCh. 19 - Prob. 7PCh. 19 - Prob. 8PCh. 19 - Prob. 9PCh. 19 - Prob. 10P
Ch. 19 - Prob. 11PCh. 19 - Prob. 12PCh. 19 - Prob. 13PCh. 19 - Prob. 14PCh. 19 - Prob. 15PCh. 19 - Prob. 16PCh. 19 - Prob. 17PCh. 19 - Prob. 18PCh. 19 - Prob. 19PCh. 19 - Prob. 20PCh. 19 - Prob. 21PCh. 19 - Prob. 22PCh. 19 - Prob. 23PCh. 19 - Prob. 24PCh. 19 - Prob. 25PCh. 19 - Prob. 26PCh. 19 - Prob. 27PCh. 19 - Prob. 28PCh. 19 - Prob. 29PCh. 19 - Prob. 30PCh. 19 - Prob. 31PCh. 19 - Prob. 32PCh. 19 - Prob. 33PCh. 19 - Prob. 34PCh. 19 - Prob. 35PCh. 19 - Prob. 36PCh. 19 - Prob. 37PCh. 19 - Prob. 38PCh. 19 - Prob. 39PCh. 19 - Prob. 40PCh. 19 - Prob. 41PCh. 19 - Prob. 42PCh. 19 - Prob. 43PCh. 19 - Prob. 44PCh. 19 - Prob. 45PCh. 19 - Prob. 46PCh. 19 - Prob. 47PCh. 19 - Prob. 48PCh. 19 - Prob. 49PCh. 19 - Prob. 50PCh. 19 - Prob. 51PCh. 19 - Prob. 52PCh. 19 - Prob. 53PCh. 19 - Prob. 54PCh. 19 - Prob. 55PCh. 19 - Prob. 56PCh. 19 - Prob. 57PCh. 19 - Prob. 58PCh. 19 - Prob. 59PCh. 19 - Prob. 60PCh. 19 - Prob. 61PCh. 19 - Prob. 62PCh. 19 - Prob. 63PCh. 19 - Prob. 64PCh. 19 - Prob. 65PCh. 19 - Prob. 66PCh. 19 - Prob. 67PCh. 19 - Prob. 68PCh. 19 - Prob. 69PCh. 19 - Prob. 70PCh. 19 - Prob. 71PCh. 19 - Prob. 72PCh. 19 - Prob. 73PCh. 19 - Prob. 74PCh. 19 - Prob. 75PCh. 19 - Prob. 76PCh. 19 - Prob. 77PCh. 19 - Prob. 78PCh. 19 - Prob. 79PCh. 19 - Prob. 80PCh. 19 - Prob. 81PCh. 19 - Prob. 82PCh. 19 - Prob. 83PCh. 19 - Prob. 84PCh. 19 - Prob. 85PCh. 19 - Prob. 86PCh. 19 - Prob. 87PCh. 19 - Prob. 88PCh. 19 - Prob. 89P
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- The gasoline internal combustion engine operates in a cycle consisting of six parts. Four of these parts involve, among other things, friction, heat exchange through finite temperature differences, and accelerations of the piston; it is irreversible. Nevertheless, it is represented by the ideal reversible Otto cycle, which is illustrated below. The working substance of the cycle is assumed to be air. The six steps of the Otto cycle ale as follows: i. Isobaric intake stroke (OA). A mixture of gasoline and air is drawn into the combustion chamber at atmospheric pressure P0 as the piston expands, increasing the volume of the cylinder from zero to VA . ii. Adiabatic compression stroke (AB). The temperature of the mixture rises as the piston compresses it adiabatically from a volume VA to VB . iii. Ignition at constant volume (BC). The mixture is ignited by a spark. The combustion happens so fast that there is essentially no motion of the piston. During this process, the added heat Q1 causes the pressure to increase from pB to pc at the constant volume VB(=Vc) . iv. Adiabatic expansion (CD). The heated mixture of gasoline and air expands against the piston, increasing the volume from VC to VD . This is called the power stroke, as it is the part of the cycle that delivers most of the power to the crankshaft. v. Constant-volume exhaust (DA). When the exhaust valve opens, some of the combustion products escape. There is almost no movement of the piston during this part of the cycle, so the volume remains constant at VA(=VD) . Most of the available energy is lost here, as represented by the heat exhaust Q2 . vi. Isobaric compression (AO). The exhaust valve remains open, and the compression from VA to zero drives out the remaining combustion products. (a). Using (i)e=W/Q1; (ii)w=Q1Q2; and (iii)Q1=nCv(TCTB),Q2=nCv(TDTA), Show that e=1TDTATCTB. (b). Use the fact that steps (ii) and (iv) are adiabatic to show that e=11r1 where r=VA/VB . The quantity r is called the compression ratio of the engine. (c) In practice, r is kept less than around 7. For larger values, the gasoline-air mixture is compressed to temperatures so high that it explodes before the finely timed spark is delivered. This preignition causes engine knock and loss of power. Show that for r=6 and =1.4 (the value for air), e=0.51 , or an efficiency of 51%. Because of the many irreversible processes, an actual internal combustion engine has an efficiency much less than this ideal value. A typical efficiency for a tuned engine is about 25% to 30%.arrow_forwardThe energy output of a heat pump is greater than the energy used to operate the pump. Why doesn't this statement violate the first law of thermodynamics?arrow_forwardCalculate the net work output of a heat engine following path ABCDA as shown below.arrow_forward
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