ANSWERS TO TEST YOURSELF QUESTIONS 3 1physics for the iB Diploma © camBriDge University press 2015

answers to test yourself questionstopic 33.1 thermal concepts

  1  a  The thermal energy lost by one body must equal the thermal energy gained by the other because of energy conservation.

  b  The changes in temperature are not, however, necessarily equal because the masses and specific heat capacities may differ.

  2  a  From the definition, Q mc cQ

m= ⇒ = =

×= − −∆

∆θ

θ385

0 150 5 00513 1 1

. .J kg K .

  b  It is the same.

  3  The energy provided is 20 3 0 60 3600× × =. J. Hence 0 090 420 4 0 0 310 4 0 3600 2 8 103. . . . .× × + × × = ⇒ = ×c c J kkg K− −1 1

.

  4  The energy provided is 40 4 0 60 9600× × =. J. Hence

25 15 8 0 140 15 8 9600 4 2 103 1× + × × = ⇒ = × −. . . .c c J kg K−−1. The obvious assumptions are that the liquid and

the calorimeter are heated uniformly and that none of the energy supplied gets lost to the surroundings.

  5  The loss of potential energy is mgh = × × = ×1360 10 86 1 17 10. 6 J. Then,

C∆ ∆θ θ= × ⇒ =×

×=1 17 10

1 17 10

16 1073

3..6

6

K

  6  a C m c m c= + = × + × = × −1 1 2 2

545 0 450 23 0 4200 1 17 10. . . J K 11.

  b  ∆ ∆∆∆

∆∆

Q CQ

tC

t= ⇒ =θ θ

. Hence 450 1 17 10 3 9 105 3 1= × × ⇒ = × − −. .∆∆

∆∆

θ θt t

K s . For a change of

temperature of 20.0 K we then require a time of 20

3 9 105 2 10 873

3

..

×= × =− s min.

  7  The energy transferred from the water and the aluminum container is Q = 0.300 × 4200 × 10 + 0.150 × 900 × 10 = 13950 J. This is used to (a) raise the temperature of ice to the melting point of 0 °C, (b) melt the ice at 0 °C and (c) raise the temperature of the melted ice (which is now water) to the final temperature of 0 °C. Thus 13950 = × × + × × + × ×m m m2200 10 334 10 4200 103 . Hence, m = 0 035. kg.

  8  The mass of ice is m = × × =20 0 06 900 1080. kg. So we need Q = × × + × × = ×1080 2200 5 1080 334 10 3 7 103 8. J.

  9  a  Let the surface area (in square meters) of the pond be A. Then in time t the energy falling on the surface will be Q A t= × ×600 . The volume of ice is V A= × 0 01. and so its mass is m A= × ×( . )0 01 900 . Then 600 0 01 900 334 103× × = × × × ×A t A( . ) . We see that the unknown surface of area cancels out and is not

required. Then, t =× × ×

= ≈0 01 900 334 10

6005010 84

3.mins .

  b  This assumes that none of the incident radiation is reflected from the ice and that all the ice is uniformly heated.

10  a Q1 J= × × = ×1 0 2200 10 2 2 104. .

  b Q23 51 0 334 10 3 34 10= × × = ×. . J

  c Q3 J= × × = ×1 0 4200 10 4 2 104. .  d  In the melting stage.

2 physics for the iB Diploma © camBriDge University press 2015ANSWERS TO TEST YOURSELF QUESTIONS 3

11  The water will lose an amount of thermal energy 1 00 4200 10 42000. × × = J. This energy is used to melt the ice and then raise the temperature of the melted ice to 10 C° . Thus m m m× × + × × = ⇒ =334 10 4200 10 42000 0 1123 . kg .

12  Since the specific latent heat of vaporisation of water is so much larger than the specific latent heat of fusion we expect that the final temperature will be greater than the initial 30 °C of the water. Then:0 150 4200 30 0 100 334 10 0 100 42003. ( ) . .× × − + × × + × ×T TT

T

= × × + ×× −

0 050 2257 10 0 050

4200 100

3. .

( ). .  This long equation can be solved for T (preferably using the Solver of your calculator) to give T = °95 C.

3.2 modelling a gas

13  There are 28

214=  moles of hydrogen and so 14 6 02 10 8 1023 24× × ≈ ×.  molecules.

14  There are 6 0

4 01 5

.

..=  moles.

15 2 0 10

6 02 103 3

24

23

.

..

××

≈  moles.

16  Krypton has 84

214 0= . moles; 4.0 moles of carbon correspond to

12

4 03 0

..= g of carbon.

17  From PV

nT

PV

n T1 1

1 1

2 2

2 2

= we deduce that P

T

P

T1

1

2

2

= i.e. That 12 0 10

295 393

52. ×

=P

hence P2516 0 10= ×. Pa. (Notice the

change to kelvin.)

18  From PV

nT

PV

n T1 1

1 1

2 2

2 2

= we deduce that PV PV1 1 2 2= i.e. That 8 2 10 2 3 10 4 5 106 3 62. . .× × × = × ×− V hence

V23 34 2 10= × −. m .

19  A quantity of 12.0 kg of helium corresponds to 12 10

43 0 10

33×

= ×. mol. Then from the gas law, pV nRT= we

get PnRT

V= =

× × ××

= ×−

3 00 10 8 31 293

5 00 101 46 10

3

39. .

.. Paa.

20  From the gas law, pV nRT= we get nPV

RT= =

× × × ××

=−4 00 1 013 10 12 0 10

8 31 2931 9

5 3. . .

.. 998 mol. Since the mass of

one mole of carbon dioxide (CO2 ) is 44 g, we need 44 1 9998 87 9× =. . g.

21  We use PV

nT

PV

n T1 1

1 1

2 2

2 2

= to get P

n

P

n1 2

2= and hence nn

21

2= . In other words to reduce the pressure to half its original

value, half the molecules must leave the container. The original number of molecules can be found using

pV nRT= to get nPV

RT= =

× × ××

=−5 00 10 300 10

8 31 3000 0602

5 6.

.. and hence N = × × = ×0 0602 6 02 10 3 62 1023 22. . . .

So we will have to lose N

21 81 1022= ×. molecules. This will take

1 81 10

3 00 10603 10

22

19

.

.min

××

= ≈s .

ANSWERS TO TEST YOURSELF QUESTIONS 3 3physics for the iB Diploma © camBriDge University press 2015

22  P

V

b

ac

23  Let there be n1 moles of the gas in the left container and n2 in the right. Then it must be true using nPV

RT=

that

nRT1

5 312 10 6 0 10=

× × × −. and n

RT1

5 36 0 10 3 0 10=

× × × −. .. When the gases mix we will have n n1 2+ moles in a volume of

9.0 dm3 and so n nP

RT1 2

39 0 10+ =

× × −.. Hence

12 10 6 0 10 6 0 10 3 0 10 9 05 3 5 3× × ×+

× × ×=

× ×− −. . . .

RT RT

P 110 3−

RT.

This means that P =× × + × ×

= × =12 10 6 0 6 0 10 3 0

9 010 10 10

5 55. . .

.Pa atm.

24  a   The cross sectional area of the piston is AV

h= = =

0 050

0 5000 10 2.

.. m . The pressure in the gas is constant and

equal to PF

A= =

×= ×

10 0 10

0 0101 0 104.

.. Pa.

  b  From the gas law, nPV

RT= =

× ××

=1 0 10 0 050

8 31 2920 206

4. .

.. . The number of molecules is then

N = × × = ×0 206 6 02 10 1 24 1023 23. . . .

  c  From V

T

V

T1

1

2

2

= we get 0 050

292 4252.

=V

hence V22 37 3 10= × −. m .

25  The mass is just 28 2 56× = g . The volume is found from VnRT

P= =

× ××

=2 0 8 31 273

1 0 100 045

53. .

.. m .

26  The molar mass of helium is 4.00 g per mole. A mass of 70.0 kg of helium corresponds to

70 0 10

4 001 75 10

34.

..

×= × mol. Thus P

nRT

V= =

× × ×= ×

1 75 10 8 31 290

4041 04 10

45. .

. Pa.

27  a nPV

nRT= =

× × ××

= ×−

−150 10 5 0 10

8 31 3003 01 10

3 42.

.. mool.

  b N nNA= = × × × = ×−3 01 10 6 02 10 1 8 102 23 22. . .

  c  M n= = × × =−µ 3 01 10 29 0 872. . g

28  a VnRT

P= =

× ××

= × −1 0 8 31 273

1 0 102 27 10

52 3. .

.. m

  b  We have 1 mole and so 4.00 g of helium. The density is thus ρ = =××

=−

−−M

V

4 00 10

2 27 100 176

3

23.

.. kg m .

  c  The change for oxygen is just the molar mass and so ρ = × = −32

40 176 1 41 3. . kg m .

4 physics for the iB Diploma © camBriDge University press 2015ANSWERS TO TEST YOURSELF QUESTIONS 3

29  Under the given changes the volume will stay the same and so the density will be unchanged.

30  We use 1

2

3

22mc kT= . The mass of a molecule is

4 0

6 02 106 64 10 6 64 10

2323 27.

.. .

×= × = ×− −g kg. Hence

ckT

m= =

× × ××

≈−

−−3 3 1 38 10 850

6 64 102300

23

271.

.m s .

31  From 1

2

3

22mc kT= we get c

kT

m=

3. The mass of a molecule (in kg) is

M

NA

4 0

6 02 106 64 10 6 64 10

2323 27.

.. .

×= × = ×− −g kg. Hence c

kT

m

kN T

M

RT

M= = =

3 3 3A since kR

N=

A

.

32  a 1

2

3

2

3

21 38 10 300 6 2 102 23 21mc kT= = × × × = ×− −. . J

  b 1

2

1

2

32

481 1

22 2

2 1

2

2

1

2

1

m c m cc

c

m

m

N

N= ⇒ = = = =

µµ

A

A