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*…show more content…*

But

Work done by motor in time

1=

(power)( l)

= (OA

hp x 746 W I hp)(l )

Equating this to our previous value for the work done yield s

I

20.7 III

=

1.26 x 10 5 J

(OA x 746 ) W

= 420 s = 7 min

In each of the following situations, find the change in internal energy of the system. (a) A system absorbs 500 cal of heat and at the same time does 400 J of work . (b) A system

CHAP. 20]

209

FIRST LAW OF THERMODYNA1"fICS

absorbs 300 cal and at the same time 420 J of work is done on it. (c) Twelve hundred calories is removed from a gas held at constant volume. Give your answers in kilojoules.

(a)

!:::"U = 2,Q - !:::"IY

(b)

= (500 cal)(4.184 llcal) -

4001 = 1.69 kl

b"U = !:::"Q -!:::"W = (300 cal)(4.184 l l cal) - (-420 1)

(e)

!:::,,[j

= b"Q -

!:::"W

= (- 1200 cal )(4.184 l/cal) - 0

=

1.68 kl

= -5.02 kJ

Notice that !:::"Q is positive when heat is added to the system. and!:::" W is positive when the system does work. In the reverse cases. b"Q and b" W mllst be taken negative.

20.8 II]

For each of the following adiabatic processes. find the change in internal energy. (a) A gas does 5 .r of work while expanding adiabatically. (b) During an adiabatic compression, 80 J of work is

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

But

Work done by motor in time

1=

(power)( l)

= (OA

hp x 746 W I hp)(l )

Equating this to our previous value for the work done yield s

I

20.7 III

=

1.26 x 10 5 J

(OA x 746 ) W

= 420 s = 7 min

In each of the following situations, find the change in internal energy of the system. (a) A system absorbs 500 cal of heat and at the same time does 400 J of work . (b) A system

CHAP. 20]

209

FIRST LAW OF THERMODYNA1"fICS

absorbs 300 cal and at the same time 420 J of work is done on it. (c) Twelve hundred calories is removed from a gas held at constant volume. Give your answers in kilojoules.

(a)

!:::"U = 2,Q - !:::"IY

(b)

= (500 cal)(4.184 llcal) -

4001 = 1.69 kl

b"U = !:::"Q -!:::"W = (300 cal)(4.184 l l cal) - (-420 1)

(e)

!:::,,[j

= b"Q -

!:::"W

= (- 1200 cal )(4.184 l/cal) - 0

=

1.68 kl

= -5.02 kJ

Notice that !:::"Q is positive when heat is added to the system. and!:::" W is positive when the system does work. In the reverse cases. b"Q and b" W mllst be taken negative.

20.8 II]

For each of the following adiabatic processes. find the change in internal energy. (a) A gas does 5 .r of work while expanding adiabatically. (b) During an adiabatic compression, 80 J of work is

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