Physics: First Law of Thermodynamics Assignment Solutions
CHAP. 20]
207
FIRST LAW OF THERMODYNAMICS
The Carnot cycle is the most efficient cycle possible for a heat engine. An engine that operates accordance to this cycle between a hot reservoir (Til) and a cold reservoir (Te) has efficiency
ef ' lmax =
In
Tc
I - -T
II
Kelvin temperatures must be used in this equation.
Solved Problems
20.1 III
In a certain process, 8.00 kcal of heat is furnished to the system while the system does
6.00 kJ of work. By how much does the internal energy of the system change during the process? We have
tlQ
= (8000
eal )(4.184 1leal)
Therefore, from the First Law tlQ
tlU
20.2 III
= 33 .5
k1
tlW
and
= 6.00 kJ
= tlU + tl w.
= tlQ -
tlW
= 33.5
kJ - 6.00 k1
= 27.5
k1
The specific heat of water is 4184 J /kg· K. By how many joules does the internal energy of
50 g of water change as it is heated from 21 °C to 37 °C? Assume that the expansion of the water is negligible.
The heat added to raise the temperature of the water is
tlQ
= em
= (4184 1/kg·K)(0.050
tlT
kg)(16 c C)
= 3.4 x
103 J
Notice that tlT in Celsius is equal 10 tl Tin kel\ 'ins. If we ignore the slight expansion of the water, no work was done on the surroundings and so tl W = O. Then. the first law, tlQ = 6,.U + tl W . tells liS thaI 6,.[:
20.3 III
= tlQ
= 3.4 kJ
How much does the internal energy of 5.0 g of ice at precisely 0 cC increase as it is changed to water at 0 C Neglect the change in volume.
C?
The heat needed to melt the ice is
6,.Q
= mLr = (5.0 g)( 80 eal/g) = 400 eal
No external work is done by the iee as it melts and so tl W
= tlU + tlW , tells us that
= O.
Therefore, the First Law,
tlQ
6,.U
20.4 IIII
= tlQ = (400 eal )(4.l84 1l eal) =
1.7 kJ
A spring (k = 500 N / m) supports a 400-g mass which is immersed in 900 g of water. The specific heat of the mass is 450 J / kg. K. The spring is now
207
FIRST LAW OF THERMODYNAMICS
The Carnot cycle is the most efficient cycle possible for a heat engine. An engine that operates accordance to this cycle between a hot reservoir (Til) and a cold reservoir (Te) has efficiency
ef ' lmax =
In
Tc
I - -T
II
Kelvin temperatures must be used in this equation.
Solved Problems
20.1 III
In a certain process, 8.00 kcal of heat is furnished to the system while the system does
6.00 kJ of work. By how much does the internal energy of the system change during the process? We have
tlQ
= (8000
eal )(4.184 1leal)
Therefore, from the First Law tlQ
tlU
20.2 III
= 33 .5
k1
tlW
and
= 6.00 kJ
= tlU + tl w.
= tlQ -
tlW
= 33.5
kJ - 6.00 k1
= 27.5
k1
The specific heat of water is 4184 J /kg· K. By how many joules does the internal energy of
50 g of water change as it is heated from 21 °C to 37 °C? Assume that the expansion of the water is negligible.
The heat added to raise the temperature of the water is
tlQ
= em
= (4184 1/kg·K)(0.050
tlT
kg)(16 c C)
= 3.4 x
103 J
Notice that tlT in Celsius is equal 10 tl Tin kel\ 'ins. If we ignore the slight expansion of the water, no work was done on the surroundings and so tl W = O. Then. the first law, tlQ = 6,.U + tl W . tells liS thaI 6,.[:
20.3 III
= tlQ
= 3.4 kJ
How much does the internal energy of 5.0 g of ice at precisely 0 cC increase as it is changed to water at 0 C Neglect the change in volume.
C?
The heat needed to melt the ice is
6,.Q
= mLr = (5.0 g)( 80 eal/g) = 400 eal
No external work is done by the iee as it melts and so tl W
= tlU + tlW , tells us that
= O.
Therefore, the First Law,
tlQ
6,.U
20.4 IIII
= tlQ = (400 eal )(4.l84 1l eal) =
1.7 kJ
A spring (k = 500 N / m) supports a 400-g mass which is immersed in 900 g of water. The specific heat of the mass is 450 J / kg. K. The spring is now