Unformatted text preview: ∆T = Q / Q = C p n ∆T = 5 R n ∆T . Thus,
2 .
c Rn h Using these expressions for ∆V and ∆T, we find that
– nR ∆T / P – nR Q / bRn g – Q
s=
=
=
5
2 5
2 A = b – –2093 J
5
2 5
2 PA g 1
c.01 × 10 Pa h3.14 × 10
c
5 –2 m2 PA =
h 0 . 264 m 94. REASONING According to the conservation of energy, the work W done by the electrical
energy is W = QH − QC , where QH is the magnitude of the heat delivered to the outside
(the hot reservoir) and QC is the magnitude of the heat removed from the house (the cold
reservoir). Dividing both sides of this relation by the time t, we have W
t = QH
t − QC
t The term W / t is the magnitude of the work per second that must be done by the electrical
energy, and the terms QH / t and QC / t are, respectively, the magnitude of the heat per second delivered to the outside and removed from the house. Since the air conditioner is a
Carnot air conditioner, we know that QH / QC is equal to the ratio TH / TC of the Kelvin
temperatures of the hot and cold reservoirs, QH / QC = TH / TC (Equation 15.14). This
expression, along with the one above for W / t , will allow us to determine the magnitude of
the work per second done by the electrical energy. SOLUTION Solving the expression QH / QC = TH / TC for QH , substituting the result
into the relation W
t = QH
t − QC
t , and recognizing that QC / t = 10 500 J/s, give Chapter 15 Problems 825 QC TH
W
t = QH
t − QC
t = TC
t − QC
t Q T 306.15 K = C H − 1 = (10 500 J/s ) − 1 = 5.0 ×102 J/s t TC 292.15 K In this result we have used the fact that TH = 273.15 + 33.0 °C = 306.15 K and TC = 273.15 +
19.0 °C = 292.15 K. Pressure (×105 Pa) 95. REASONING
a. The work done by the gas is equal to the area under the pressureversusvolume curve.
We will measure this area by using the graph given with the problem. 6.00
4.00 B A 2.00
0
0 2.00 4.00 6.00 8.00 10.0 12.0 Volume, m3 b. Since the gas is an ideal gas, it obeys the ideal gas law, PV = nRT (Equation 14.1). This
implies that PAVA / TA = PBVB / TB . All the variables except for TB in this relation are
known. Therefore, we can use this expression to find the temperature at point B.
c. The heat Q that has been added to or removed from the gas can be obtained from the first
law of thermodynamics, Q = ∆U + W (Equation 15.1), where ∆U is the change in the
internal energy of the gas and W is the work done by the gas. The work W is known from
part (a) of the problem. The change ∆U in the internal energy of the gas can be obtained
from Equation 14.7, ∆U = U B − U A = 3 nR (TB − TA ) , where n is the number of moles, R is
2
the universal gas constant, and TB and TA are the final and initial Kelvin temperatures. We
do not know n, but we can use the ideal gas law (PV = nRT ) to replace nRTB by PBVB and
to replace nRTA by PAVA. 826 THERMODYNAMICS SOLUTION
a. From the drawing we see that the area under the curve is 5.00 “squares,” where each ( )( ) square has an area of 2.00 × 105 Pa 2.00 m3 = 4.00 × 105 J. Therefore, the work W done
by the gas is ( ) W = ( 5.00 squares ) 4.00 ×105 J/square = 2.00 × 106 J b. In the Reasoning section, we have seen that PAVA / TA = PBVB / TB . Solving this relation
for the temperature TB at point B, using the fact that PA = PB (see the graph), and taking the
values for VB and VA from the graph, we have that
PV
TB = B B
P V AA V 10.0 m3 TA = B TA = (185 K ) = 925 K V 2.00 m3 A c. From the Reasoning section we know that the heat Q that has been added to or removed
from the gas is given by Q = ∆U + W. The change ∆U in the internal energy of the gas is
∆U = U B − U A = 3 nR (TB − TA ) . Thus, the heat can be expressed as
2
Q = ∆U + W = 3 nR (TB − TA ) + W
2 We now use the ideal gas law (PV = nRT ) to replace nRTB by PBVB and nRTA by PAVA.
The result is
3
Q = 2 ( PBVB − PAVA ) + W
Taking the values for PB, VB, PA, and VA from the graph and using the result from part a
6 that W = 2.00 × 10 J, we find that the heat is
Q= 3
2 ( PBVB − PAVA ) + W ( )( )( )( ) = 3 2.00 ×105 Pa 10.0 m3 − 2.00 ×105 Pa 2.00 m3 + 2.00 ×106 J = 4.40 ×106 J
2 96. REASONING We think of the entire universe as divided into two parts: the sun and the rest
of the universe. As the sun spontaneously radiates heat into space, the change in its entropy
each second is ∆Ssun , and the change in the entropy of the rest of the universe is ∆Srest each
second. The total entropy change of the universe in one second is the sum of the two entropy
changes
∆Suniverse = ∆Ssun + ∆Srest
(1) Chapter 15 Problems 827 The spontaneous thermal radiation of the sun is an irreversible process, but if we imagine a
reversible process by which the sun loses an amount Q of heat, and a reversible process by
Q
which the rest of the universe absorbs this heat, then we can make use of ∆S = T R
(Equation 15.18) to calculate the entropy changes of the sun and of the rest of the universe:
∆Ssun = −Q
Tsun ∆S rest = and +Q
Trest (2) In Equations (2), Tsun is the sun’s surface temperature, and Tre...
View
Full
Document
This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details