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Date November 2017 Marks available 3 Reference code 17N.2.SL.TZ0.4
Level Standard level Paper Paper 2 Time zone 0 - no time zone
Command term Show that Question number 4 Adapted from N/A

Question

A large cube is formed from ice. A light ray is incident from a vacuum at an angle of 46˚ to the normal on one surface of the cube. The light ray is parallel to the plane of one of the sides of the cube. The angle of refraction inside the cube is 33˚.

Each side of the ice cube is 0.75 m in length. The initial temperature of the ice cube is –20 °C.

Calculate the speed of light inside the ice cube.

[2]
a.i.

Show that no light emerges from side AB.

[3]
a.ii.

Sketch, on the diagram, the subsequent path of the light ray.

[2]
a.iii.

Determine the energy required to melt all of the ice from –20 °C to water at a temperature of 0 °C.

Specific latent heat of fusion of ice  = 330 kJ kg–1
Specific heat capacity of ice            = 2.1 kJ kg–1 k–1
Density of ice                                = 920 kg m–3

[4]
b.i.

Outline the difference between the molecular structure of a solid and a liquid.

[1]
b.ii.

Markscheme

«v = c sin  i sin  r  =»  3 × 10 8 × sin ( 33 ) sin ( 46 )

2.3 x 108 «m s–1»

a.i.

light strikes AB at an angle of 57°

critical angle is «sin–1 ( 2.3 3 ) =» 50.1°

49.2° from unrounded value

angle of incidence is greater than critical angle so total internal reflection

OR

light strikes AB at an angle of 57°

calculation showing sin of “refracted angle” = 1.1

statement that since 1.1>1 the angle does not exist and the light does not emerge

[Max 3 marks]

a.ii.

total internal reflection shown

ray emerges at opposite face to incidence

Judge angle of incidence=angle of reflection by eye or accept correctly labelled angles

With sensible refraction in correct direction

a.iii.

mass = «volume x density» (0.75)3 x 920 «= 388 kg»

energy required to raise temperature = 388 x 2100 x 20 «= 1.63 x 107

energy required to melt = 388 x 330 x 103 «= 1.28 x 108 J»

1.4 x 108 «J» OR 1.4 x 105 «kJ»

Accept any consistent units

Award [3 max] for answer which uses density as 1000 kg–3 (1.5× 108 «J»)

b.i.

in solid state, nearest neighbour molecules cannot exchange places/have fixed positions/are closer to each other/have regular pattern/have stronger forces of attraction

in liquid, bonds between molecules can be broken and re-form

OWTTE

Accept converse argument for liquids

[Max 1 Mark]

b.ii.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
a.iii.
[N/A]
b.i.
[N/A]
b.ii.

Syllabus sections

Core » Topic 4: Waves » 4.4 – Wave behaviour
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