Monday, October 26, 2015

Physics 9702 Doubts | Help Page 218

  • Physics 9702 Doubts | Help Page 218

Question 1040: [Applications > Remote Sensing > Magnetic Resonance]
Outline briefly the main principles of the use of magnetic resonance to obtain information about internal body structures.

Reference: Past Exam Paper – June 2008 Paper 4 Q10

Solution 1040:
A large / strong (constant) magnetic field causes the nuclei to rotate about the direction of the field / precess.
An applied radio frequency / r.f. pulse causes resonance in the nuclei, and nuclei absorb energy. The (pulse) is at the Larmor frequency.
On relaxation / the nuclei de-excite emitting (a pulse at) r.f. This pulse is detected and processed.
A non-uniform field (superimposed) allows for the position of the nuclei to be determined and for the location of detection to be changed.

Question 1041: [Matter > Elastic and Plastic Behaviour > Young modulus]
In stress-strain experiments on metal wires, the stress axis is often marked in units of 108 Pa and the strain axis is marked as a percentage. This is shown for a particular wire in the diagram.

What is the value of the Young modulus for the material of the wire?
A 6.0 × 107 Pa             B 7.5 × 108 Pa             C 1.5 × 109 Pa             D 6.0 × 109 Pa

Reference: Past Exam Paper – June 2010 Paper 11 Q19 & Paper 12 Q21 & Paper 13 Q20

Solution 1041:
Answer: D.
Young modulus = Stress / Strain

From the stress-strain graph, the Young modulus can be obtained by the gradient.
Consider the points (5, 3×108) and (0, 0)
Note that a strain of 5% is equal to 0.05.

Gradient = (3×108 – 0) / (0.05 – 0) = 6.0 × 109 Pa

Question 1042: [Matter > Ideal Gases]
(a) (i) State what is meant by the internal energy of a system.
(ii) Explain why, for an ideal gas, the internal energy is equal to the total kinetic energy of the molecules of the gas.

(b) The mean kinetic energy <EK> of a molecule of an ideal gas is given by the expression
<EK> = (3/2) kT
where k is the Boltzmann constant and T is the thermodynamic temperature of the gas.
A cylinder contains 1.0 mol of an ideal gas. The gas is heated so that its temperature changes from 280 K to 460 K.
(i) Calculate the change in total kinetic energy of the gas molecules.
(ii) During the heating, the gas expands, doing 1.5 × 103 J of work.
State the first law of thermodynamics. Use the law and your answer in (i) to determine the total energy supplied to the gas.    

Reference: Past Exam Paper – November 2013 Paper 43 Q2

Solution 1042:
(i) The internal energy of a system is the sum of kinetic and potential energies of the molecules due to their random motion.

(ii) For an ideal gas, there are no intermolecular forces. So there is no potential energy (only kinetic).

{Mean kinetic energy <EK> of a molecules of an ideal gas at temperature T = (3/2) kT
1 mole of an ideal gas contains 1.0 × 6.02×1023 molecules
Change in KE of a molecule = (3/2) kΔT = (3/2) k (460 – 280) = (3/2) k (180)}
change in kinetic energy = 3/2 × 1.38×10–23 × 1.0 × 6.02×1023 × 180
change in kinetic energy = 2240 J

R = kNA
{So, k = R / NA
For 1 mole, <EK> = (3/2) kT × NA
Since k = R / NA: <EK> = (3/2) (R/NA) T × NA = (3/2) RT}
energy = 3/2 × 1.0 × 8.31 × 180
energy = 2240 J

1st law of thermodynamics:
increase in internal energy = heat supplied + work done on system
{As explained in (a)(ii), for an ideal gas, the internal energy = total kinetic energy. Since the gas is doing work, the energy is taken with a negative sign.}
2240 = energy supplied – 1500
energy supplied = 3740 J

Question 1043: [Matter > Thermal properties of materials]
(a) Define specific latent heat.

(b) An electrical heater is immersed in some melting ice that is contained in a funnel, as shown in Fig. 3.1.

The heater is switched on and, when the ice is melting at a constant rate, the mass m of ice melted in 5.0 minutes is noted, together with the power P of the heater. The power P of the heater is then increased. A new reading for the mass m of ice melted in 5.0 minutes is recorded when the ice is melting at a constant rate.
Data for the power P and the mass m are shown in Fig. 3.2.

(i) Complete Fig. 3.2 to determine the mass melted per second for each power of the heater.

(ii) Use the data in the completed Fig. 3.2 to determine
1. a value for the specific latent heat of fusion L of ice,
2. the rate h of thermal energy gained by the ice from the surroundings.

Reference: Past Exam Paper – June 2015 Paper 41 & 43 Q3

Solution 1043:
(a) Specific latent heat is (numerically equal to) the quantity of (thermal) energy/heat to change state/phase of unit mass at constant temperature

{5.0 minutes = 5.0×60 s}
at 70 W, mass s–1 = {78 / (5.0×60) =} 0.26 g s–1
at 110 W, mass s–1 = {114 / (5.0×60) =} 0.38 g s–1   

{Amount of heat energy = mL
Divide the above by time        (Energy / time = Power)
Power = (mass/time) × L
(mass/time) is the mass melted per second, as calculated for the last column above
But we need to account for the energy gains from the surroundings. This is done by using the difference in powers and the differences in masses for both set of values in the table.
Let the rate of thermal energy gained by the ice from the surroundings be h.
Both the heat gained from the surroundings and the heater would contribute in melting the ice.}
P + h = mL     
{By using both sets of values, we are eliminating ‘h’ the value of which is unknown yet. Actually this is a simultaneous equation. Consider the equation: P + h = mL
Replace the 1st set of values from the table: 100 + h = 0.38L            …..(1)
Replace the 2nd set of values from the table: 70 + h = 0.26L             …..(2)
Take the difference of the 2 equations to eliminate h}
(110 – 70) = (0.38 – 0.26) L
Specific latent heat of fusion L of ice = 330 J g–1

{Now the L is known, we can replace either set of values from the table to obtain h.}
EITHER 70 + h = 0.26 × 330             OR 110 + h = 0.38 × 330
h = 17 / 16 / 15 W


  1. Please consider answering ALL of the following questions:
    4/O/N/02 Q.5(b),Q.6(c)(i)
    6/O/N/02 Q.11(a)(b)
    6/O/N/03 Q.9
    04/M/J/04 Q.8(a),(b)(i),(ii)1.
    06/M/J/04 Q.9(b)(iii),Q.11(b)
    06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)
    04/M/J/05 Q.7(a)
    06/O/N/05 Q.8(b),Q.10(a)
    04/M/J/06 Q.6(a),(c),Q.7(b)
    06/M/J/06 Q.14(b)
    04/O/N/06 Q.3(c)
    06/O/N/06 Q.3(b)
    05/M/J/07 Q.2(d)
    04/O/N/07 Q.10(c)
    04/M/J/08 Q.5(b),Q.9(b)
    51/M/J/10 Q.2(d)

    1. For 04/M/J/06 Q.6(a),(c), see solution 1046 at

  2. Hi, do you have a page like this for chemistry? I really need it for chemistry heh. thank you in advance :)

    1. Check at

      There's not much though

  3. question 1042, since mean KE= 3/2kT, why did you times with the no. of moles?

    1. As state din the question, this is for A molecule.
      1 mole contains 1.0 × 6.02×10^23 molecules


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