• Physics 9702 Doubts | Help Page 116

Question 588: [Temperature > Internal energy]

(a) On Fig.1, place tick (✓) against those changes where internal energy of the body is increasing.

water freezing at constant temperature           ..........................................

a stone falling under gravity in a vacuum       ..........................................

water evaporating at constant temperature     ..........................................

stretching a wire at constant temperature       ..........................................

(b) A jeweller wishes to harden a sample of pure gold by mixing it with some silver so that mixture contains 5.0% silver by weight. Jeweller melts some pure gold and then adds the correct weight of silver. Initial temperature of the silver is 27 °C. Use data of Fig.2 to calculate initial temperature of the pure gold so that the final mixture is at the melting point of pure gold.

gold                 silver

melting point / K                                                                     1340                1240

specific heat capacity (solid or liquid) / J kg^–1K^–1                   129                  235

specific latent heat of fusion / kJ kg^–1                                     628                  105

(c) Suggest a suitable thermometer for measurement of the initial temperature of the gold in (b).

Reference: Past Exam Paper – June 2003 Paper 4 Q2

Solution 588:

(a)

water freezing at constant temperature           ..........................................

a stone falling under gravity in a vacuum       ..........................................

water evaporating at constant temperature                  tick

stretching a wire at constant temperature                    tick

{The concept of ‘Internal Energy’

Internal energy is the sum of potential and kinetic energy due to the RANDOM motion of the molecules.

It is important to point out that the internal energy is due to the energies of the randomly moving molecules only AND as the word says internal energy is due to the ‘internal’ structure of the system (due to molecules or atoms), not the whole system itself.

For example, for a car moving with speed v or moving up a slope. The sum of its kinetic energy and gain in potential energy do NOT give the internal energy. It only gives the total energy of the car.

So, again when we are talking about internal energy, we are firstly considering the atoms / molecules inside the materials and secondly, we are only considering those atoms / molecules that are in random motion.}

{Case 1: water freezing at constant temperature

For constant temperature, the kinetic energy of the molecules is unchanged. For freezing, there is a change in phase from liquid to solid occurring. This requires energy to form the new bonds. So, the material is actually losing some of its potential energy to form the bonds. So, internal energy is decreasing in this case.

Case 2: a stone falling under gravity in a vacuum

This case is dealing with the energy of a large object, not its internal structure. As explained before, in such a case, we do not talk about internal energy, but rather the total energy of the stone.

Case 3: water evaporating at constant temperature

Again for constant temperature, the kinetic energy of the molecules is unchanged. For evaporation, there is a change in phase from liquid to gas occurring. The potential energy between the randomly moving molecules as they go in the gas state (more molecules are now in random motion). So, internal energy is increasing in this case.

Case 4: stretching a wire at constant temperature

Again for constant temperature, the kinetic energy of the molecules is unchanged. For stretching a wire, work needs to be done on the wire. Since the wire has now become longer than its initial length, the molecules / atoms inside the wire have more space to move freely / randomly compared to its initial shape where all the molecules / atoms were packed together (in the solid wire). We say that the wire has elastic potential energy. So, the internal energy of the wire increases.}

(b)

{To reach the final temperature which is equal to the melting point of pure gold, the temperature of the silver (initially at 300K) should increase. This means that the temperature of the pure gold should have decreased to provide energy to the silver.

ΔQ = mcΔθ

m is the total mass. 95% of m is gold and 5% of m is silver.}

Heat lost by liquid gold = 0.95m × 129 × ΔT

{The melting point go gold is higher than that of silver. So, before reaching the final temperature, the silver has melted first. This requires energy which is given by ΔQ = mL = 0.05m × 105 000 where L is the specific latent heat of fusion.

After melting, the silver needs to increase its temperature to the final temperature. This further requires energy given by ΔQ = mcΔθ = 0.05m × 235 × (1340 – 300)}

Heat gained (by silver) = 0.05m × 235 × (1340 – 300) + 0.05m × 105 000

{Heat lost by liquid gold = heat gained by silver}

122.5mΔT = 17 470m

Change in temperature ΔT = 143 K

{Final temperature is the melting point of gold}

Initial temperature of gold = 143 + 1340 = 1483 K

(c) Example: thermocouple / resistance thermometer

Question 589: [Current of Electricity]

Four resistors of equal value are connected as shown.

How will powers to the resistors change when resistor W is removed?

A The powers to X, Y and Z will all increase.

B The power to X will decrease and the powers to Y and Z will increase.

C The power to X will increase and the powers to Y and Z will decrease.

D The power to X will increase and the powers to Y and Z will remain unaltered.

Reference: Past Exam Paper – June 2011 Paper 11 Q36

Solution 589:

Answer: C.

Power dissipated in a resistor = I²R

Connecting 2 resistors of, say, resistance R in series causes the overall resistance to increase (= R+R = 2R).

Connecting 2 resistors of, say, resistance R in parallel causes the overall resistance to decrease (= [1/R + 1/R]^-1 = 0.5R).

Since the 4 resistors have the same resistance in the connection shown, the p.d. across each parallel connection (and each resistor) will be the same.

Overall resistance in circuit = 0.5R + 0.5R = R

Let the e.m.f. in the circuit = E

Current in the circuit = E / R = I

The total current splits at the parallel combination.

Power dissipated in X = (I/2)²R = I²R / 4 = 0.25 I²R

The same power is dissipated in all the resistors.

When W is removed, the overall resistance of the parallel combination of W and X changes from 0.5R to R (due to X only).

Total resistance in new circuit = R + 0.5R = 1.5R

Current in circuit = E / 1.5R = (2/3)I

The overall current in the circuit will also change – it decreases since the overall resistance in the complete circuit has increased. This current flows through X but is split into half at the combination of Y and Z. So, the current through Y and Z is half that through X.

Power dissipated = I²R

Power dissipated in X = [(2/3)I]²R = 4 I²R / 9 = 0.44 I²R

Half the current flows through Y and Z. The same power is dissipated through both resistors.

Power dissipated in Y = [0.5 (2/3)I]²R = I²R / 9 = 0.11 I²R

So, the power in X increases and the power in Y (and Z) decreases.

Question 590: [Electromagnetism > Hall Probe]

(a) Constant current is maintained in a long straight vertical wire. Hall probe is positioned a distance r from the centre of the wire, as shown in Fig.1.

(i) Explain why, when Hall probe is rotated about the horizontal axis XY, the Hall voltage varies between a maximum positive value and a maximum negative value.

(ii) Maximum Hall voltage V_H is measured at different distances r.

Data for V_H and the corresponding values of r are shown in Fig.2.

V_H / V             r / cm

0.290               1.0

0.190               1.5

0.140               2.0

0.097               3.0

0.073               4.0

0.060               5.0

It is thought that V_H and r are related by expression of the form

V_H = k / r

where k is a constant.

1. Without drawing a graph, use data from Fig.2 to suggest whether the expression is valid.

2. A graph showing variation with 1 / r of V_H is plotted.

State features of the graph that suggest that the expression is valid.

(b) Hall probe in (a) is now replaced with a small coil of wire connected to sensitive voltmeter. Coil is arranged so that its plane is normal to the magnetic field of the wire.

(i) State Faraday’s law of electromagnetic induction and hence explain why voltmeter indicates a zero reading.

(ii) State three different ways in which an e.m.f. may be induced in the coil.

Reference: Past Exam Paper – June 2010 Paper 41 Q5

Solution 590:

(a)

(i) The Hall voltage V_H depends on the angle between (the plane of) the probe and B-field (magnetic field).

either The Hall voltage V_H is maximum when the plane and the B-field are normal to each other.

or The Hall voltage V_H is zero when the plane and the B-field are parallel

or The Hall voltage V_H depends on the sine of the angle between the plane and B-field

(ii)

1.

{From the formula, the product of VH and r is a constant. k = V_H r}

Calculate V_H r at least three times {any sets can be chosen}

1^st set: 0.29 × 1.0 = 0.29 (to 2sf) = 0.3

2^nd set: 0.19 × 1.5 = 0.285 = 0.29 (to 2sf) = 0.3

3^rd set: 0.14 × 2.0 = 0.28 (to 2sf) = 0.3

EITHER If given to 1 s.f., the product is constant, so the expression is valid or approximately constant so valid

OR If given to 2 s.f., the product is not constant, so the expression is invalid

2. The graph is a straight line that passes through the origin

(b)

(i) Faraday’s law of electromagnetic induction states that the e.m.f. induced is proportional / equal to the rate of change of (magnetic) flux (linkage).

The voltmeter indicates a zero reading because there is a constant field in coil / flux (linkage) of coil does not change

(ii) Example:

Vary the current (in wire) / switch current on or off / use a.c. current

Rotate the coil

Move the coil towards / away from wire

Question 591: [Work, Energy and Power]

Box of weight 30 N is released from rest on a ramp that is at angle of 30° to the horizontal. The box slides down the ramp so that it falls through a vertical distance of 8.0 m. A constant frictional force of 10 N acts on box while it is moving.

What is the kinetic energy of box after falling through this distance?

A 80 J                          B 160 J                        C 240 J                        D 400 J

Reference: Past Exam Paper – November 2014 Paper 13 Q17

Solution 591:

Answer: A.

In this question, there is a frictional force of 10N that opposes the motion of the box down the ramp. So, work needs to be done against friction.

Consider a case where the ramp was frictionless. From the conservation of energy, the gravitational potential energy would be converted into kinetic energy of the box as it falls.

But in this case, friction is present. So, not all of the gravitational potential energy would be kinetic energy. Some of the energy is used up as work done against friction.

Loss in GPE = Gain in KE + Work against friction

Loss in GPE = mgh = (mg)h = (30) (8) = 240 J

Let the distance travelled by the box along the ramp = s

sin30 = 8 / s

Distance s = 8 / sin30 = 16m

Work done against friction = Fs = 10 (16) = 160 J

Loss in GPE = Gain in KE + Work against friction

240 = Gain in KE + 160

Gain in KE = 240 – 160 = 80 J