• Physics 9702 Doubts | Help Page 97

Question 500: [Thermodynamics]

(a) Resistance of thermistor at 0^oC is 3840Ω. At 100^oC resistance is 190Ω. When thermistor is placed in water at a particular constant temperature, resistance is 2300Ω.

(i) Assuming that resistance of thermistor varies linearly with temperature, calculate the temperature of water

(ii) Temperature of water, as measured on thermodynamic scale of temperature, is 286K. By reference to what is meant by thermodynamic scale of temperature, comment on answer in (i)

(b) Polystyrene cup contains mass of 95g of water at 28^oC.

Cube of ice of mass 12g is put into water. Initially, ice is at 0^oC. The water, of specific heat capacity 4.2x10³Jkg^-1K^-1, stirred until all ice melts.

Assuming that cup has negligible mass and that there is no heat exchange with atmosphere, calculate final temperature of water

(Specific latent heat of fusion of ice is 3.3x10⁵Jkg^-1)

Reference: Past Exam Paper – June 2010 Paper 42 & 43 Q3



Solution 500:

(a)

(i)

A change of 1^oC corresponds to (3840 – 190) / 100 Ω (= 36.5Ω)

So, for a resistance of 2300Ω,

Temperature = 100 x [(2300 - 3840) / (190 - 3840)] = 42^oC.

(ii)

EITHER 286K = 13^oC           OR 42^oC = 315K

The thermodynamic scale of temperature does not depend on the property of a substance. So, the change in resistance (of the thermistor) with temperature is non-linear.

(b)

{Heat gained by the ice in melting = mass x specific latent heat of fusion of ice}

Heat gained by the ice in melting = 0.012 x 3.3x10⁵ = 3960J

Heat lost by the water (= mcΔθ) = 0.095 (4.2x10³) (28 – θ)

{Heat energy of MELTED ice to reach temperature θ = 3960 + [0.012 (4.2x10³) (θ – 0)])

The heat lost by the water is used by the ice to reach temperature θ}

3960 + [0.012 (4.2x10³) (θ – 0)] = 0.095 (4.2x10³) (28 – θ)

Temperature θ = 16^oC

Question 501: [Thermistor]

Variation with temperature of resistance R_T of thermistor shown in Fig. 

Thermistor connected in circuit of Fig. 

Battery has e.m.f. 9.0V and negligible internal resistance. Voltmeter has infinite resistance.

(a) For the thermistor at 22.5^oC, calculate

(i) total resistance between points A and B on Fig

(ii) reading on the voltmeter

(b) Temperature of the thermistor is changed. Voltmeter now reads 4.0V.

Determine

(i) total resistance between A and B on Fig

(ii) Temperature of the thermistor

(c) Student suggests that the voltmeter, reading up to 10V, could be calibrated to measure temperature.

Suggest 2 disadvantages of using the circuit of Fig with this voltmeter for measurement of temperature in range 0^oC to 25^oC

Reference: Past Exam Paper – November 2010 Paper 22 Q6



Solution 501:
Go to
The variation with temperature of the resistance RT of a thermistor is shown in Fig. 6.1.

Question 502: [Current of Electricity]
Diagram shows an arrangement of resistors.

What is total electrical resistance between X and Y?
A less than 1 Ω
B between 1 Ω and 10 Ω
C between 10 Ω and 30 Ω
D 40 Ω

Reference: Past Exam Paper – November 2010 Paper 11 Q36



Solution 502:
Answer: B.
The arrangement represents a parallel combination of a 10 Ω with a set of three 10 Ω resistors that are connected in series.

Total resistance of the set of three resistors = 10 + 10 + 10 = 30 Ω

Let the equivalent resistance across X and Y be R.
1 / R = 1/10 + 1/30
Equivalent resistance R = [1/10 + 1/30]^-1 = 7.5 Ω










Question 503: [Waves > Diffraction grating]
Parallel beam of red light of wavelength 700 nm is incident normally on diffraction grating that has 400 lines per millimetre.
What is the total number of transmitted maxima?
A 3                              B 4                              C 6                              D 7

Reference: Past Exam Paper – June 2013 Paper 13 Q27



Solution 503:
Answer: D.
Diffraction grating: d sinθ = nλ
Order of maxima, n = d sinθ / λ

Slit separation, d = 1x10^-3 / 400 = 2.5x10^-6 m
Wavelength, λ = 700x10^-9 m

The greatest angle (if possible) for a maxima to form is at 90^o.
Order of maxima, n = (2.5x10^-6) (sin90^o) / (700x10^-9) = 3.57 = 3
Since the order n can only be an integer, the greatest number of maxima on one side is 3.

But diffraction occurs on both sides of the incident light, so there are 3 maxima on each side of the incident light. Additionally, one maxima is (undeflected) obtained at the centre, where the light is incident.

Total number of maxima = 3 + 3 + 1 = 7