# 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 10

^{8}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 × 10

^{7}Pa B 7.5 × 10^{8}Pa C 1.5 × 10^{9}Pa D 6.0 × 10^{9}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, 3×10

^{8}) and (0, 0)
Note that a strain of 5% is equal to 0.05.

Gradient = (3×10

^{8}– 0) / (0.05 – 0) = 6.0 × 10^{9 }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 <E

_{K}> of a molecule of an ideal gas is given by the expression

<E

_{K}> = (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 × 10

^{3}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:__**(a)**

(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).

**(b)**

(i)

EITHER

{Mean
kinetic energy <E

_{K}> of a molecules of an ideal gas at temperature T = (3/2) kT
1 mole
of an ideal gas contains 1.0 × 6.02×10

^{23}molecules
Change
in KE of a molecule = (3/2) kÎ”T = (3/2) k (460 – 280) = (3/2) k (180)}

change in
kinetic energy = 3/2 × 1.38×10

^{–23}× 1.0 × 6.02×10^{23}× 180
change in
kinetic energy = 2240 J

OR

R = kN

_{A}
{So,
k = R / N

_{A}
For 1
mole, <E

_{K}> = (3/2) kT × N_{A}
Since
k = R / N

_{A}: <_{EK}> = (3/2) (R/N_{A}) T × N_{A}= (3/2) RT}
energy = 3/2 ×
1.0 × 8.31 × 180

energy = 2240 J

(ii)

1

^{st}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

**(b)**

(i)

{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}
(ii)

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 1

^{st}set of values from the table: 100 + h = 0.38L …..(1)
Replace the 2

^{nd}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}
2.

{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

Please consider answering ALL of the following questions:

ReplyDelete4/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)

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

Deletehttp://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-219.html

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

ReplyDeleteCheck at

Deletehttp://chemistry-ref.blogspot.com/

There's not much though

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

ReplyDeleteAs state din the question, this is for A molecule.

Delete1 mole contains 1.0 × 6.02×10^23 molecules