7/30/2019 Physics Ans

1/5

1. C

2. B

3. B

4. B

5. C

6.

D

7. A

8. B

9. A

10. D

11. D

12. B

13. B

14. B

15. D

16. C

17. B

18. B

19. C

20. A

21. D22. D

23. B

24. C

25. D

26. C

27. D

28. C

29. D

30. D

7/30/2019 Physics Ans

2/5

31. (a) In one minute, 2 kg of water flows through the heater.

Energy supplied by heater = energy absorbed by water

Pt= mcT

8000 60 = 2 4200 (T 15)

T= 72.1 C

The temperature of the hot water is 72.1 C.

1M

1M

1A

(b) In one minute,

energy lost by hot water = energy gained by cold water

m1cT1= m2cT2

2 (72.1 42) = m2 (42 15)

m2= 2.23 kg

The flow rate of cold water is 2.23 kg min1.

1M

1A

32. (a) TA= 30 C

Let mbe the initial mass of water in each cup.

Energy gained by cold water = energy lost by hot water

1A

1M

2

1mc(TB 30) = mc(60 TB) 1M

2

1TB 15 = 60 TB

TB= 50 C 1A

(b) TA will be unchanged.

TBwill be lower than 50 C.

1A

1A

33. (a) Energy absorbed = Pt

= 1000 180

= 180 000 J

1M

1A

(b) By Q= mcT, 1M

specific heat capacity =Tm

Q

=)50200(5.0

000180

= 2400 J kg1C1 1A

(c) Since the temperature of the pot increases, it absorbs energy from the immersion

heater.

The energy actually absorbed by the cooking oil is less than that supplied by the

heater.

Hence, the actual value of the specific heat capacity of the cooking oil is lowerthan that found in (b).

1A

1A

1A

7/30/2019 Physics Ans

3/5

34. (a) After some time, the water inside and outside the aluminium can is heated to 100

C.

The energy transfer between the water inside and outside the can is zero.

Therefore, the water inside the can cannot obtain latent heat of vaporization.

1A

1A

1A

(b) Power =t

Q 1M

=t

Tmc 1M

=90

)25100(420015.0

= 525 W 1A

(c) Energy transferred = mlv

= 0.005 2.26 106

= 11 300 J

1M

1A

35. (a) Energy lost by horseshoe = energy absorbed by water

mHcT= mWcT+ mWlv

0.5 450 (500 100) = mW [4200 (100 20) + 2.26 106]

1M

1M

m= 34.7 g

The maximum amount of water vaporized is 34.7 g.

1A

(b) Less water will be vaporized

because more energy from the horseshoe will be used to raise the temperature ofthe whole bucket of water.

1A

1A

36. (a) Energy required to heat 5 bowls of water from 20 C to 100 C

= mcT

= (0.15 5) 4200 (100 20)

= 252 000 J

Energy required to vaporize 2 bowls of water

= mlv

= (0.15 x 2) x 2.26 x 106 J

= 678 000 J

Total energy required = 252 000 + 678 000

= 930 000 J

1M

1M

1M

1A

(b) Some energy is lost to the surroundings during heating. 1A

37. (a) The water in flask A takes the shortest time to drop 1 C.

The dull black surface of the flask is a good radiator of heat.

The large temperature difference between the water and the surroundings leadsto a high rate of heat loss.

1A

1A

1A

(b) The rate of heat loss of water decreases. 1A

7/30/2019 Physics Ans

4/5

38. (a) Put the thermometer in boiling water and mark the mercury level.

Then put the thermometer in melting ice and mark the mercury level.

The temperatures of the boiling water and melting ice are taken as 0 C and 100

C respectively.

The separation between these two mercury levels is divided into 100 equal

divisions and each division is 1 C.

1A

1A

1A

1A

(b) By proportion,

room temperature =0.40.24

0.44.8

100 1M

= 22 C 1A

(c) (i) Initial temperature = 22 C

Final temperature = 0.40.24

0.45.15

100 = 57.5C 1M

Energy supplied by heater = energy absorbed by liquid L

Pt= mcT

50 5 60 = 0.5 c (57.5 22)

c= 845 J kg1C1

1M

1M

1A

The specific heat capacity of liquid L is 845 J kg1C1.

(ii) The result of experiment is different from the actual value because energy is

lost to the surroundings. 1AMethods to reduce heat loss to the surroundings (any two of the following):

Use a more powerful heater to reduce the heating time.

Replace the glass with a polystyrene cup.

Add a lid to cover the cup.

Add a polystyrene tile under the cup.

2

1A

39. (a)

(Heater and ice placed in the funnel)

(Beaker placed under the funnel)

(Power supply, joulemeter and heater connected correctly)

1A

1A

1A

7/30/2019 Physics Ans

5/5

(b) He should use melting ice. 1A

(c) Energy supplied by heater = 46 400 - 35 000 = 11 400 J 1M

Specific latent heat of fusion of ice =m

Q

1M

=04.0

40011

= 2.85 x 105 J kg1 1A

(d) To measure the amount ice melted at room temperature, a set of control

apparatus is needed. 1A

The control apparatus is identical to that of the experimental apparatus except that

the immersion heater is not turned on. 1A

Before putting the beakers underneath the funnels and switching on the heater in

the experimental apparatus, make sure that the drip rates are about the same.

1A

After switching off the heater, do not remove the beakers immediately. Wait until

the drip rates are about the same again. 1A

(e) Ice can be prepared easily.

Ice has a large specific latent heat of fusion.

It melts at room temperature.

1A

1A

1A

40. (a) Metal is a better conductor of heat than paper.

It conducts energy away from the hand more easily.

1A

1A

(b) (i) Radiation 1A

(ii) The candle flame heats up the air around it. 1A

The hot air expands and rises,

producing a convection current.

1A

1A

The air flowing upwards pushes the fan of the cylinder and drives the cylinder

to rotate. 1A

(iii) The plastic bag will rise. 1A

(iv) The cylinder will rotate.

This is because the dry ice cools the air.

Cold air sinks and pushes the fan of the cylinder.

1A

1A

1A