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Unformatted text preview: find that c hc h Q = ∆U + W = −4.2 × 10 4 J + 1.6 × 10 4 J = −2 .6 × 10 4 J 2. REASONING In both cases, the internal energy ∆U that a player loses before becoming
exhausted and leaving the game is given by the first law of thermodynamics: ∆U = Q − W
(Equation 15.1), where Q is the heat lost and W is the work done while playing the game.
The algebraic signs of the internal energy change ∆U and the heat loss Q are negative, while
that of the work W is positive.
SOLUTION
a. From the first law of thermodynamics, the work W that the first player does before
leaving the game is equal to the heat lost minus the internal energy loss: ( ) W = Q − ∆U = −6.8 ×105 J − −8.0 ×105 J = +1.2 ×105 J b. Again employing the first law of thermodynamics, we find that the heat loss Q
experienced by the morewarmlydressed player is the sum of the work done and the
internal energy change: ( ) Q = W + ∆U = 2.1×105 J + −8.0 ×105 J = −5.9 ×105 J
As expected, the algebraic sign of the heat loss is negative. The magnitude of the heat loss
is, therefore, 5.9 ×105 J . Chapter 15 Problems 3. 767 SSM REASONING Energy in the form of work leaves the system, while energy in the
form of heat enters. More energy leaves than enters, so we expect the internal energy of the
system to decrease, that is, we expect the change ∆U in the internal energy to be negative.
The first law of thermodynamics will confirm our expectation. As far as the environment is
concerned, we note that when the system loses energy, the environment gains it, and when
the system gains energy the environment loses it. Therefore, the change in the internal
energy of the environment must be opposite to that of the system.
SOLUTION
a. The system gains heat so Q is positive, according to our convention. The system does
work, so W is also positive, according to our convention. Applying the first law of
thermodynamics from Equation 15.1, we find for the system that b gb g ∆U = Q − W = 77 J − 164 J = −87 J As expected, this value is negative, indicating a decrease.
b. The change in the internal energy of the environment is opposite to that of the system, so
that ∆U environment = +87 J . 4. REASONING The first law of thermodynamics, which is a statement of the conservation
of energy, states that, due to heat Q and work W, the internal energy of the system changes
by an amount ∆U according to ∆U = Q − W (Equation 15.1). This law can be used directly
to find ∆U.
SOLUTION Q is positive (+7.6 × 104 J) since heat flows into the system; W is also positive
(+7.6 × 104 J) since work is done by the system. Using the first law of thermodynamics from
Equation 15.1, we obtain ( )( ) ∆U = Q − W = +7.6 ×104 J − +4.8 ×104 J = +2.8 ×104 J
The plus sign indicates that the internal energy of the system increases. 5. SSM REASONING Since the change in the internal energy and the heat released in the
process are given, the first law of thermodynamics (Equation 15.1) can be used to find the
work done. Since we are told how much work is required to make the car go one mile, we
can determine how far the car can travel. When the gasoline burns, its internal energy
decreases and heat flows into the surroundings; therefore, both ∆U and Q are negative. 768 THERMODYNAMICS SOLUTION According to the first law of thermodynamics, the work that is done when one
gallon of gasoline is burned in the engine is
W = Q − ∆U = −1.00 × 10 8 J – (–1.19 × 10 8 J) = 0.19 × 10 8 J Since 6.0 × 10 5 J of work is required to make the car go one mile, the car can travel
0.19 × 10 8 J 6. F 1 mile I =
G × 10 J J
H
K
6.0
5 32 miles REASONING According to the discussion in Section 14.3, the internal energy U of a
monatomic ideal gas is given by U = 3 nRT (Equation 14.7), where n is the number of
2
moles, R is the universal gas constant, and T is the Kelvin temperature. When the
temperature changes to a final value of Tf from an initial value of Ti, the internal energy
changes by an amount
U f − U i = 3 nR (Tf − Ti )
124 2
43
∆U
2
Solving this equation for the final temperature yields Tf = ∆U + Ti . We are given n 3nR and Ti , but must determine ∆U. The change ∆U in the internal energy of the gas is related to
the heat Q and the work W by the first law of thermodynamics, ∆U = Q − W
(Equation 15.1). Using these two relations will allow us to find the final temperature of the
gas.
SOLUTION Substituting ∆U = Q − W into the expression for the final temperature gives 2
Tf = ( Q − W ) + Ti 3nR 2 = +2438 J − ( −962 J ) + 345 K = 436 K 3 ( 3.00 mol ) 8.31 J/ ( mol ⋅ K ) Note that the heat is positive (Q = +2438 J) since the system (the gas) gains heat, and the
work is negative (W = −962 J), since it is done on the system. Chapter 15 Problems 7. 769 REASONING AND SOLUTION
a. For the weight lifter
∆U = Q − W
= −mLv − W = −(0.150 kg)(2.42 × 106 J/kg) − 1.40 × 105 J = −5.03 × 105 J
b. Since 1 nutritional calorie = 4186 J, the numb...
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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

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