Exam Style Answers 19 Asal Physics CB

 

 

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded may be different.

 Coursebook answers Chapter 19 This is clearly a nonsense! We can see how fundamental this law is, and why it is called the zeroth law.  [2]

Exam-style questions 1

C 

[1]

2

B 

[1]

3

a

Kinetic energy remains constant. 

[1]

 Potential energy increases.  Potential

[1]

 Internal energy increases Internal increases.. 

[1]

b

Kinetic energy increases increases.. 

[1]

 

Potential energy remains constant (slight increase/decrease if water expands/ contracts).  [1]

  4

Internal energy increases increases.. 

7

 

q heat (energy) transferred to the system W  work  work done on system  

 b

i

 ii ii

[1]

Just before the stone hits the ground, it has kinetic energy. energy. All the molecules are moving together in the same direction.  [1]

The air is being compressed, so work is being done on it. 

  8

 IIn n this case, there is no energy supplied by heating and little is lost, but work is done in compressing the air air..  [1] If two bodies are at the same temperature, no energy flows from one body to another. Therefore,, if no energy flows from A to B and Therefore none flows from B to C, but energy flowed from C to B, it would mean that A and B were at the same temperature, and A and C are at

[1]

[1]

work done on gas W  = p  p∆V = 4 × 105  × 3 × 10−4 = + 120 J  [1] =− + =−   U     220 120   100 J No work done on or by the gas so W  = 0  ∆

heat is removed from the gas; the temperature falls 

[1] [1]

−330 = q so

a

No temperature difference, so 0 V  

b

temperature = 

 

= 77.8 ≈ 78

c

i

voltage = 

 

= 206

 ii ii

That the variation between

100 × 49 63

 

°C 

[1] [1] [1] [1]

63 × 327 100

µV 

µV 

[1] [1]

temperature difference and the e.m.f. e.m.f. in the thermocouple remains linear beyond 100 °C  [1]

[1]

From  F rom the first law of thermodynamics t hermodynamics,, the change in internal energy of a body is equal to the energy supplied by heating plus the energy supplied by doing work.  [1]

increase in internal energy 

 

 

The  T he kinetic energy for movement in random directions is internal energy, and hence the temperature rises.  [1]

6

∆U  an  an

[1]

When  W hen it hits the ground, this unidirectional movement of the molecules is converted into movement of individual molecules in random directions.   [1]

5

a

9

a

1 °C corresponds to a change of 18 Ω   [1]  temperature = (620 − 200)/18 = 23 °C  [1] temperature

b

280 K = 7° C or 23° C = 296 K 

[1]

 

Thermodynamic scale does not depend on the property of a substance.  [1]

 

Change in resistance is not linear with temperature.   [1]

10 a

energy supplied = mc∆θ = 0.300 ×  4180 × 80 

[1]

energy supplied = 100 320 J 

[1]

 

the same temperature, but C is hotter than B.

1

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020

   

CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK 

energy

100320

13 a  

[1]

 

time = 

 

= 201

 

No energy is needed to heat the element or the kettle  [1]

b

energy supplied = power × time energy

        11 a

power

=

500

[1]

s

supplied = 500 × 120 = 60 000 J   [1] energy mass of water boiled aw away ay =  s.l.h. 60000 =  = 0.027 kg  [1] 2.2 26  6 × 106 mass remaining = 300 − 27 = 273 ≈ 270 g   [1] No energy is lost to the surroundings; all the vapour escapes from the kettle.  [1] The energy required to raise the temperature of unit mass of a substance   [1]

 

by unit temperature rise 

b

i

heat needed = 0.020 × 2100 × 15 +  0.020 × 330 000  [1]

 

= 7230

 ii  ii

heat lost by water in cooling =  0.200 × 4200 × (26 − T ) 

 

  12 a

b

2

 

as they are moved further apart (accept bonds broken) / work is done pushing back the atmosphere  [1]

b

To reduce the energy gained from the surroundings 

[1]

 c

energy input = 40 × 2 × 60 (= 4800 J) 

[1]

 

Use of E  = ml  

[1]

 

m = mass decrease with heater on minus half mass decrease with heater off   [1]

 

m = 23.8 g 4800 l  =   = 202 J g−1  23.8

  14 a

J 

[1]

random distribution (of kinetic energy) 

[1]

Lowest temperature (at which energy cannot be removed from molecules) 

[1]

 

0 K 

[1]

i

mass per second =  ρ ρAv

 

mass per second = 1000 × 4.8 × 10−5 × 1.2 

[1]

 

mass per second = 0.058 kg 

[1]

 ii ii

E  = mc∆θ leading to

   ii ii

[1] [1]

heat gained by ice in melting and then warming to temperature T = 7230 +  0.02 × 4200 × T   [1] T = 15.8 or 16 °C  

[1]

b

The energy needed to change the state of unit mass mass of a substance substance  [1] [1]

Latent heat of fusion is heat needed to change form solid to liquid and latent heat of vaporisation is heat needed to change the state from liquid to gas.  [1] i 

Each minute the mass decreases by the same amount.  [1]

 ii  ii

energy provided = 120 × 60 = 7200 J [1]

 

L = 7200/0.0062 = 1.2 × 106 J kg−1  [1]

 iii iii

Too large 

 

Heat lost from beaker means less than 7200 J is used to boil the liquid.   [1]

[1]

Sum of the kinetic energy and potential energy of the molecules

i

[1]

 without change in temperature  without  

Energy goes to potential energy of the molecules  [1]

∆θ = 

9000 0.05  8 × 4200

 

°C 

 

= 37

 

final temperature of the water = 37 + 15 = 52 °C 

[1] [1] [1]

 iii iii

The heater is 100% efficient; no heat is lost or gained from the pipe.  [1]

 iv iv

Decrease the rate of flow of water  [1]

[1]

Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside © Cambridge University Press 2020