Category: Physic Tutorials

How to solve questions on current of electricity for UTME

Question 1

A working electric motor takes a current of 1.5A when the p.d. across it is 250V. If its efficiency is 80%, the power output is{UTME 2001}
A. 469.0W        B. 300.0W        C. 4.8W          D. 133.0W

Solution

Efficiency = (power output / power input)%

power input = IV = 1.5*250 = 375W

80/100 = power output / 375

power output = 80*375 / 100 = 300 W

B is the correct option

Question 2

A bread toaster uses a current of 4A when plugged in a 240 volts line. It takes one minute to toast slices of bread. What is the energy consumed by the toaster?{UTME 2001}
A. 3.60 x 103J       B. 5.76 x 104J      C. 1.60 x 102J        D. 1.60 x 104J

Solution

one minute is 60 seconds

Energy consumed = IVt = 4*240*60 = 57600 J = 5.76 x 104 J

B is the correct answer

Question 3

In the circuit diagram above, the ammeter reads a current of 3A when R is 5Ω and 6A when R is 2Ω. Determine the value of x.{UTME 2001}
A. 10Ω            B. 8Ω         C. 4Ω            D. 2Ω

Solution

X and R are in parallel

Effective resistance = (X*R)/(R+X)

When current is 3A, the effective resistance = 5X /(5+X)

V = IR

V = 3[5X /(5+X)]

when current is 6A, the effective resistance = 2X /(2+X)

V = IR

V = 6[2X /(2+X)]

Since R and X are in parallel, the same voltage will be applied across the two resistors

3[5X /(5+X)] = 6[2X /(2+X)]

15X /(5+X) = 12X/(2+X)

15/(5+X) = 12/(2+X)

cross multiply

15(2+x) = 12(5+x)

30 +15x = 60 + 12x

3x = 30

x = 10Ω

A is the correct answer

Question 4

A cell of emf 1.5V is connected in series with a 1Ω resistor and a current of 0.3A flows through the resistor. Find the
internal resistance of the cell.{UTME 2013)
A. 1.0 Ω        B. 4.0 Ω          C. 3.0 Ω            D. 1.5 Ω

Solution

E = I(R+r)

1.5 = 0.3(1 + r)

1 + r = 1.5/0.3

1 + r =5

r = 5 – 1 = 4Ω

B is the correct option

Question 5

Which of the following obeys ohms law?{UTME 2013}
A. Glass             B. Glectrolytes            C. Metals                D. Diode

Solution

Ohm’s law is about metallic material only provided the temperature and all other physical factors rmain constant.

C is the correct option

Question 6

Six identical cells, each of e.m.f. 2V are connected as shown above. The effective e.m.f. of the cell is{UTME 2012}
A. 0V        B. 4V        C. 6V          D. 12V

Solution

cells arranged in series will have effective emf of

E = E1 + E2 + E3 +…

E = 2+2+2+2+2+2 = 12 V

D is the correct option

Question 7

In the circuit above, three resistors, 2Ω, 4 Ω and 12 Ω are connected in parallel and a 12 V battery is connected across the combination. The current flowing through the 12 Ω resistor is{UTME 2011}
A. 9.6 A             B 14.4 A               C. 1.0 A               D. 3.2 A.

Solution

The current flowing through the 12 Ω resistor will be, V =IR

Since they are all in parallel, they will have the same voltage across them,

12 = I*12

I = 12/12 = 1A

C is the correct option

Question 8

The diagram above shows a balanced metre bridge, the value of x is{UTME 2010}
A. 66.7 cm              B. 25.0 cm              C. 33.3 cm                 D. 75.0 cm.

Solution

R1 /R2 = L1 / L2

R1, the resistors are in parallel, so the effective resistance = 8*8/(8+8) = 64/16 =4Ω

4 / x = 8 / (100-x)

4(100-x) = 8x

400 -4x = 8x

12x = 400

x = 400 / 12 = 33.33 cm

C is the correct option

Question 9

In the diagram above, a 200 W bulb is lighted by a 240 V a.c mains supply. If 1kWh is sold at N 40, the cost of keeping the bulb lighted for a day is{UTME 2010}

A. N 192.00              B. N 1.92                    C. N 19.20                    D. N 1,920.00.

Solution

Energy consumed = power(KW) * time(hour)

Energy consumed = 200/1000 * (a day = 24hrs) = 0.2*24 = 4.8KWh

1kWh is sold at N 40

4.8KWh will be N40*4.8 = N192.00

A is the correct option

Recommended: Questions and answers on current of electricity for Cambridge A level

How to solve questions on alternating current circuit for UTME

Question 1

The current output form of an a.c. source is given as I = 10 sin w t. The d.c. equivalent of the current is {UTME 2012}
A. 5.0A       B. 7.1A        C. 10.0A        D. 14.1A

Solution

I = I0 sin wt

Comparing this with the equation in the question

Io = maximum current = 10 A

The dc equivalent of the current = Irms

Irms = 0.7071*I0 = 0.7071 * 10 = 7.1 A

B is the correct answer

Question 2

In an a.c. circuit, the ratio of r.m.s value to peak value of current is {UTME 2011}
A. √2    B. 2       C. 1/2         D. 1/√2

Solution

I_rms = (1/√2) I₀

Irms / I0 = 1/√2

D is the correct answer

Question 3

In alternating current circuit at resonance, the angle of lead or lag is {UTME 2010}

A. π/2    B. 0     C. π/3      D. π

Solution

At resonance Xl(inductive reactance) = Xc(capacitive reactance)

The angle of lead or lag = 0

B is the correct option

Question 4

From the diagram above, if the potential difference across the resistor, capacitor and inductor are 80V, 110V and 40V respectively, the effective potential difference is{UTME 2009}
A. 116.3V        B. 50.0V         C. 230.0V      D. 106.3V

Solution

V^2 = Vr^2 + (Vc – Vl)^2

V = √(Vr^2 + (Vc – Vl)^2)

V = √(80^2 + (110-40)^2)

V = √6400 + 4900 = √11300

V = 106.3 v

D is the correct option

Question 4

The d.c. generator has essentially the same components as the a.c. generator except the presence of{UTME 2008}
A. slip-ring          B. carbon brushes         C. split ring         D. armature

Solution

The only difference between a.c and d.c generator is that d.c uses split ring which enables current to be maintained in only one direction while a.c generator uses slip-ring which enables the cureent to flow sinusoidally.

d.c has split ring

C is the correct option

Question 5

The instantaneous value of the induced e.m.f. as a function of time is ε = εo sinwt where εo is the peak value of the
e.m.f. The instantaneous value of the e.m.f, one quarter of the period, is{UTME 2007}

A. εo    B. εo/2       C.  εo/4       D. 0

Solution

w = 2πf

f = 1/T

w = 2π / T

one quarter of the period will be

w = 2π / (T/4) =4* (2π / T)

ε = εo/4

C is the correct option

Question 6

From the diagram, the inductive reactance and the resistance R are respectively{UTME 2007}
A. 10Ω and 50Ω       B. 20Ω and 50Ω          C. 25Ω and 50Ω       D. 50Ω and 45Ω

Solution

Xl = 2πFL  = 2*π*50/π * 0.1 =10Ω

V = IZ

75 = 1.5*Z

Z = 75 /1.5 = 50Ω

Z = √R^2 + Xl^2

50 = √R^2 + 10^2

50^2 = R^2 + 100

2500 = R^2 + 100

R^2 = 2400

R = √2400 = 49.98Ω  approximately = 50Ω

A is the correct option

Question 7

A transformer is rated 240V. If the primary coil is 4000 turns and the secondary voltage 12V, determine the number of turns in the secondary coil.{UTME 2005}
A. 100                   B. 150                     C. 200                   D. 250

Solution

Vp / Vs = Np / Ns

240 / 12 = 4000 / Ns

Ns = ( 4000 * 12) / 240

Ns = 200 v

C is the correct option

Recommended: Short note on alternating current

Question 1

Induced emfs are best explained using{UTME 2013}
A. Lenz’s law     B. Ohm’s law      C. Faraday’s law      D. Coulomb’s law

Solution

Athough both Faraday and Lenz explained the induced emf

Faraday’s law talk about the emf is induced as a result of the rate of change of magnetic flux linkage while

Lenz says the the induced emf move in such a way as to oppose the change producing

Look at the two theory Faraday BEST explained induced emf
C is the correct option

Question 2

A conductor of length 1m moves with a velocity of 50ms-1 at an angle of 300 to the direction of a uniform magnetic
field of flux density 1.5 Wbm-2. What is the e.m.f. induced in the conductor?{UTME 2012}
A. 37.5V        B. 50.5V          C. 75.0V                  D. 80.5V

Solution

E = Blv sinQ

E = 1.5 * 1 * 50 sin30

E = 37.5 v

A is the correct answer

Question 3

A particle carrying a charge of 1.0 x10-8C enters a magnetic field at 3.0 x102 ms-1 at right angles to the field. If
the force on this particle is 1.8 x 10-8N, what is the magnitude of the field?{UTME 2012}
A. 6.0 x 10-1T       B. 6.0 x 10-2T            C. 6.0 x 10-3T             D. 6.0 x 10-4T

Solution

F = Bqv

B = F / qv

B = 1.8 x 10-8 / (1.0 x10-8 * 3 x 102)

B = 0.6 x 10-2 = 6.0 x 10-3 T

C is the correct option

Question 4

A step-down transformer has a power output of 50W and efficiency of 80%. If the mains supply voltage is 200V,
calculate of the primary current of the transformer.{UTME 2008}
A. 0.31A                   B. 3.20A                 C. 3.40A                  D. 5.00A

Solution

efficiency = power output / power input

80/100 = 50/power input

50*100 / 80 = power input

power input = 62.5

power input = IpVp

Ip = 62.5 / 200 = 0.31 A

A is the correct option

Question 5

Two long parallel wires X and Y carry currents 3A and 5A respectively. If the force experienced per unit length by X is 5 x 10-5N, the force per unit length experienced by wire Y is{UTME 2007}
A. 3 x 10-5N             B. 3 x 10-6N                    C. 5 x 10-4N                D. 5 x 10-5N

Solution

Following the above formula and supporting it with Newton’s third law that states the if object X exert a force on object y, object y will exert the same force on X but in opposite direction.

For two parallel current carrying conductor, regardles of their current, they will exert the same force on each other but in opposite direction

D is the correct answer

Read short note on magnetic field and em induction

Notes on application Physics for Cambridge A level students

ELECTRONIC SENSOR

SENSING UNIT : This produce input voltage to processor depending on change in physical properties e.g temperatue, light or pressure.

Piezo electric Transducer

1. It converts energy from one form to another e.g microphone
2. Quartz, a crystal has its positive and negative ion joined at the centre
3. When pressure is applied the shape of the crystal changes
4. ons move apart to set the P.D which is amplify as output
5. Voltage is positive when pressure is above the ambient pressure
6. Voltage is negative when pressure is below the ambient pressure

OUTPUT DEVICES

• RELAY
• LIGHT EMMITTING DIODE
• CALIBRATED DEVICES

RELAY

• A relay is an electromagnetic switch that uses a small current to switch on or off a larger current.
• It is also used to switch on large voltages by means of small voltages
• Isolate circuit from high voltage
• Remote switching

LIGHT EMMITTING DIODE

PROCESSING UNITS

Vout = A0 (V^+ – V^–)

where A0 is the open-loop gain of the op-amp

The ideal operational amplifier (op-amp) has the following properties:

• infinite input impedance
• infinite open-loop gain
• zero output impedance
• infinite bandwidth
• infinite slew rate

MAJOR USES OF AN OPERATIONAL AMPLIFIER

1. Used as a comparator
2. Used as an inverting amplifier
3. Used as a non-inverting amplifier

NEGATIVE FEEDBACK
The process of taking some, or all, of the output of the amplifier and adding it to the input is known as feedback.

ADVANTAGES OF NEGATIVE FEEDBACK

• increased bandwidth,
• less distortion,
• greater operating stability

The circuit for an inverting amplifier is shown below

An input signal Vin is applied to the input resistor Rin. Negative feedback is applied by means of the resistor Rf. The resistors Rin and Rf act as a potential divider between the input and the output of the operational amplifier.

Assumptions
There are two basic assumptions

In order that the amplifier is not saturated, the two input voltages must be almost the same. The non-inverting input (+) is connected directly to the zero-volt line (the earth) and so it is at exactly 0 V. Thus,

the inverting input (–) must be virtually at zero volts (or earth) and for this reason, the point P is known as a virtual earth.

The input impedance of the op-amp itself is very large, therefore, there is no current in either the non-inverting or the inverting inputs. It means the current in Rf is approximately equal to the current in RIN.

Recommended: Solved questions on capacitance

Notes on quantization of energy

Max planck explained that energy from such bodies is emitted in separate or discrete packets of energy known as energy quanta or photo of amount hv where v is the frequency of radiation and he represents Planck’s constant, then energy by the equation

E = nhf

f       =        c/λ

c is the speed of light

λ is the wavelength

f is the frequency

Atomic Energy Levels

The arrangement of electrons around their nuclei is in a position known as energy levels or electron orbits of electron shells. Electrons in orbit nearest to the nucleus have the highest energy and are said to be in the ground state or lowest levels.

When an electron jumps from one level, say E4 to a lower one  E1 a photon of electromagnetic radiation is emitted with energy equal to the energy of the two levels.

hv      =        E4 – E1

f       =        c/λn

En  –  Eo = hfn     = hc/  λn

        En =  energy in the  excited state

Eo =  ground state energy

Atomic Spectra, Colour and Light Frequency

When gas atoms are executed by heating or by sending an electrical discharged, they give off light which when analyzed consist of a vast number of spectral lines. The line consists of light of one wavelength or color. This type of spectrum is called a line spectrum or the atomic spectrum of the element.

A line spectrum – Is a number of well-defined lines each having a particular frequency or wavelength or colour.

The Photoelectric Effect: This occurs when light falls on metal surfaces, electrons are emitted the emitted electrons called photoelectrons.

Eistein equation

Ek max = hf – wo

wo is the workfunction = hfo

Recommended: How to solve questions on electric field for Cambridge A level and UTME

How to solve questions on simple machine and light waves for UTME

Question 1

A ray of light strikes a plane mirror at an angle of incidence of 35o. If the mirror is rotated through 10o, through what angle is the reflected ray rotated?{utme 2001}
A. 45o          B. 20o              C. 70o                     D. 25o

The reflected ray will rotate through the angle = 2* angle through which the mirror rotate = 2*10 =20

B is the correct option

Question 2

A wheel and axle is used to raise a load of 500 N by the application of an effort of 250N. If the radii of the wheel and
the axle are 0.4cm and o.1cm respectively, the efficiency of the machine is {2002}
A. 20%                     B. 40%                  C. 50%                         D. 60%

Solution

velocity ratio = radius of wheel / radius of axle = 0.4 / 0.1 = 4

mechanical advantage = load / effort = 500 / 250 = 2

efficiency = m.a /v.r *100% = 2/4 *100% = 50%

C is the correct option

Question 3

A concave mirror of radius of curvature 40cm forms a real image twice as large as the object. The object distance is

{utme 2002}
A. 10cm                      B. 30cm                 C. 40cm                            D. 60cm.

Solution

radius = 2f

f = 40/2 = 20cm

v = image, u = object

v = 2*u = 2u

1/f = 1/u + 1/v

1/20 = 1/u + 1/2u

1/20 = 3/2u

cross mutiply

2u = 60

u = 60/2 = 30cm

B is the correct answer

Question 4

To produce an enlarged and erect image with a concave mirror, the object must be positioned{2002}
A. at the principal focus.       B. beyond the centre of curvature        C. between the principal focus and
the centre of curvature.         D. between the principal focus and the pole.

D is the correct option

Question 5

By what factor will the size of an object placed 10cm from a convex lens be increased if the image is seen on a screen placed 25cm from the lens?{utme 2003}
A. 15.0                    B. 2.5                      C. 1.5                    D. 0.4

Solution

magnification = image distance / object distance = 25 / 10 = 2.5

B is the correct option

Question 6

If an object is placed between two parallel plane mirrors with their reflecting surfaces facing each other, how many images of the object will be formed?
A. Infinite             B. Eight            C. Four                 D. Two.

Solution

N =( 360/θ) – 1

Where the angle of inclination and N is the number of images formed

angle between two parallel mirrors is 0

N = 360/0 -1 = infinity

A is the correct option

Question 7

At what position will an object be placed in front of a concave mirror in order in order to obtain an image at
infinity?{utme 2003}
A. At the pole of the mirror.      B. At the principal focus.               C. At the centre of curvature             D. Between the principal focus and the centre of curvature.

Solution

B is the correct option

Question 8

A machine whose efficiency is 60% has a velocity ratio of 5. If a force of 500N is applied to lift a load P, what is the
magnitude of P?{utme 2004}
A. 750N                 B. 4166N                  C. 500N                       D. 1500N

Solution

efficiency = m.a /v.r *100%

60/100 = m.a/5

cross multiply

m.a = 300/100 = 3

m.a = load / effort

3 = p / 500

p = 1500N

D is the correct answer
short note on optics

Notes on optics : light waves

Transmission of light

Light can travel through a solid, liquid and gas or vacuum. Therefore, it does not need a maternal medium for its transmission.

Types of object/substance

1          Opaque object: These are objects that do not allow light to pass through them, e. g wood, stone, table, wall, etc.

2         Transparent objects: Transparent objects are those which allow some light energy to pass through them, e. g plane glass, clean and clear water, etc.

3          Translucent object: This allow some light to pass through them, but object cannot be seen clearly through them, e. g some window panes, cloth.

Rectilinear Propagation of Light

Rectilinear propagation of light is the phenomenon of light travelling in a straight line. The phenomenon can be demonstrated by using a candle flame, a string and three pieces of cardboard.

The pinhole camera

This operates on the principle that light travels in a straight line. It consists of a light-tight box, one end of which has a small hole made with a pin or needle point. If the pinhole is small, the image is bright (sharp) but when it is large a brighter but blurred image is obtained. A wide hole will produce several images that overlap that are seen as single blurred image. When the object distance is far from the pinhole, it produces a invented but diminished image. Also, when the distance between the pinhole and the screen is increased, the image is enlarged but less bright.

Magnification produced by a pinhole camera = image distance / object distance or image height/ object height

Laws of Reflection

• The angle of incidence is equal the angle of reflection
• The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.

Formation of image by a plane mirror

The images formed by a plane mirror has the following characteristics

• It is erect
• It is far behind the mirror as the object is in front
• It is virtual
• It has the same size as the object
• Laterally inverted

Image formed by inclined mirror

When two mirrors are inclined at a certain angle to each other, and an object is placed in front of them, the number of images seen is given by

N =( 360/θ) – 1

Where the angle of inclination and N is the number of images formed

Uses of the plane mirror

• As looking glass
• In the periscope of two plane mirrors inclined at and fixed facing each other
• Plane mirror can be used as kaleidoscope. It consists of three plane mirrors inclined at   to each other and a piece of ground glass set at the bottom.

Curved or spherical mirror

Curved mirror are produced by cutting out a part of a spherical shell: it is called concave mirror when light is reflected from the inside and convex mirror when light is reflected from the outside.

A concave mirror can form invented, real and magnified image of the object or real, inverted and diminished  image or erect, virtual and magnified image depending on where the object is located.

• When the object is beyond C – Nature of image is real, inverted and diminished
• When the object is at C – Nature of image is real, inverted and same size
• When the object is between F and C – Nature of image is real, inverted and magnified
• When the object is at f – Nature of image is real, inverted, but formed at infinity
• When the object is between p and f – Nature of image is virtual, erect and magnified

C is the centre of curvature

F is the principal focus

p is the pole

Use of the curved mirror

• Convex mirror is used as car driving mirrors because they form upright image and also has a wild field of view.
• Convex mirror is also use in a super market to see round corners.

Concave mirrors

• Used as shaving mirror.
• Used as the distance
• Used in focusing starts for astronomical studies when used in the astronomical telescope.

Mirror formula can also be used to find the nature of the image formed by concave and convex mirror

Where V = image distance U = object distance F = focal length, F = r/2

Example

An object 8cm high placed 30cm from a concave mirror of radius of curvature 10cm. Calculate the (i) image distance (ii) image height

solution

U = 30cm  f = 20/2 = 10cm

using the mirror formula

1/f = 1/u + 1/v

1/5 = 1/30 + 1/v

1/v = 1/5 – 1/30

1/v = 6/30

v = 5 cm

image distance = 5 cm

image distance/object distance = image height / object height

5/30 = x/8

x = 40 / 30

x = 1.3 cm

image height = 1.3 cm

Refractive index

Refractive index, also called index of refraction, measure of the bending of a ray of light when passing from one medium into another. If i is the angle of incidence of a ray in vacuum (angle between the incoming ray and the perpendicular to the surface of a medium, called the normal) and r is the angle of refraction (angle between the ray in the medium and the normal), the refractive index n is defined as the ratio of the sine of the angle of incidence to the sine of the angle of refraction; i.e., n = sin i / sin r. Refractive index is also equal to the velocity of light c of a given wavelength in empty space divided by its velocity v in a substance, or n = c/v. The speed of light is not constant as it moves from medium to medium. When light enters a denser medium (like from air to glass) the speed and wavelength of the light wave decrease while the frequency stays the same. How much light slows down depends on the new medium’s index of refraction, n. (The speed of light in a medium with index n is c/n.)

UTME past questions and solutions on optics

Notes on simple machine

MACHINES

A machine is any device by means of which work can be done more conveniently.

Mechanical Advantage (M.A.)

The M.A. of a machine is defined as the ratio of load to the effort

Mechanical Advantage (M.A.) = load / effort

Velocity Ratio (V.R.)

The velocity ratio V.R. of a machine is defined as the distance moved by the effort to the distance moved by the load.

Velocity Ratio (V.R.) =    distance moved by the effort / distance moved by the load

Efficiency

The efficiency of a machine is the ratio of the useful work done by a machine to the total work put into the machine.

i.e. Efficiency = ( workout / workinput)*100%

Also, Efficiency =  (M.A /V.R)*100%

THE LEVER

A lever is a simple machine. It consists of a rigid body which is pivoted about a point called the fulcrum. The lever is based on the principle of moment. The V.R is the ratio of the two arms of the lever.

1st class Lever

The fulcrum (F) is between the load (L) and the effort (E). The velocity ratio is usually greater than 1 but could be less than or equal to 1.

Examples of First Order Lever are: See-saw, crowbar, claw hammer, pliers, a pair of scissors or pincers.

2nd Class Lever

The load is between the fulcrum and the effort. M.A and V.R are always greater than 1.

Examples of Second order lever are: nutcracker and wheelbarrow.

3rd Class Lever

 The effort is between the fulcrum and the load. M.A and V.R are less than 1.

Examples of third order lever are: forearm, forceps, sugar tongs, and table knife.

THE PULLEYS

A pulley is a wheel with a grooved rim, and there can be several of these mounted in a framework called a block. The effort is applied to a rope which passes over the pulleys.

THE BLOCK AND TACKLE

The block and tackle is the type of pulley system used in cranes and lifts. It consists of two blocks each with one or more pulleys

In a block and tickle system the V.R is  always equal to the total number of pulleys in the two blocks together.

INCLINED PLANE

A heavy load may be raised more easily by pulling it along a sloping surface than by lifting it vertically. If l is the length of the plane and h its height, then;

V.R = L /h

For a perfect inclined plane, Load x distance moved by load = Effort x distance moved by effort

In the right–angle triangle

V.R = 1 /Sin θ

THE WHEEL AND AXLE

V.R = R / r = radius of wheel / radius of axle

Recommended: Solved questions on simple machine

This article is written to help students in secondary schools and candidates writing UTME/JAMB on how to solve questions on Heat energy, specific heat capacity, and latent heat

 

How to solve questions on Heat energy (specific heat capacity)

Question 1

A 2000W electric heater is used to heat a metal object of mass 5kg initially at 10oC. If a temperature rise of 30oC is
obtained after 10 min, the heat capacity of the material is{utme 2003}
A. 6.0 x 104JoC-1             B. 4.0 x 104JoC-1             C. 1.2 x 104JoC-1                D. 8.0 x 103JoC-1

Solution

p*t = mcθ

p is the power

θ is the temperature change

the time must be converted to seconds

2000*10*60 = 5*(30-10)*c

c = 1200000/100

c = 12000JoC-1

C is the correct option

Question 2

A 50W electric heater is used to heat a metal block of mass 5kg. If in 10 minutes, a temperature rise of 12oC is
achieved, the specific heat capacity of the metal is {utme 2004}
A. 500 J kg-1 K-1                B. 130 J kg-1 K-1                 C. 390 J kg-1 K-1              D. 400 J kg-1 K-1

sol

p*t = mcθ

p is the power

θ is the temperature change

the time must be converted to second

50*10*60 = 5*12*c

c = 30000 / 60

c = 500J kg-1 K-1

A is the correct answer

Question 3

10^6J of heat is required to boil off completely 2kg of a certain liquid. Neglecting heat loss to the surroundings, the latent heat of vaporization of the liquid is {utme 2005}
A. 5.0 x 106 Jkg-1            B. 2.0 x 106 Jkg-1            C. 5.0 x 105 Jkg-1                D. 2.0 x 105 Jkg-1

Solution

Q = mL

L is the latent heat of vapourization

1000000 = 2L

L = 1000000 / 2

L  = 500000 Jkg-1

Question 4

2 kg of water is heated with a heating coil which draws 3.5A from a 200V mains for 2 minutes. What is the increase in temperature of the water? {utme 2007}
A. 25oC                      B. 15oC                            C. 10oC                     D. 30oC

Solution

Ivt = mcθ

specific heat capacity of water is 4200jkg-1k-1

convert the time to second

3.5*200*2*60 = 2*4200*θ

θ = 10^oc

Question 5

The quantity of heat energy required to melt completely 1kg of ice at -30oC is {utme 2012}
A. 4.13 x 106J          B. 4.13 x 105J            C. 3.56 x 104J           D. 3.56 x 102J
(latent heat of fusion = 3.5 x 105 Jkg-1, specific heat capacity of ice = 2.1 x 103 J kg-1 K-1)

solution

Q = mcθ + mL

Q = 1*(0-(-30))*2100 + 1*350000

Q = 63000 + 350000

Q = 413000J

B is the correct option

Question 6

Two liquids X and Y having the same mass are supplied with the same quantity of heat. If the temperature rise in X is twice that of Y, the ratio of specific heat capacity of X to that of Y is{UTME 2013}
A. 1 : 4  B. 2 : 1   C. 1 : 2   D. 4 : 1

Solution

Qx = Qy

Q = mcθ

Mx = My

mcθ(x) = mcθ(y)

θ(x) = 2θ(y)

Cx*2θ(y) = Cyθ(y)

Cx /Cy = 1 / 2

C is the correct option

Question 7

Calculate the temperature change when 500 J of heat is supplied to 100g of water.
A. 12.1oC
B. 2.1oC
C. 1.2oC
D. 0.1oC
(Specific heat capacity of water = 4200Jkg-1K-1)

Solution

Q = mcθ

100g = 0.1kg

500 = 0.1*4200* θ

θ = 500 /4200

θ = 1.2

C is the correct option

Question 8

A block of aluminium is heated electrically by a 30 W heater. If the temperature rises by 10⁰C in 5minutes, the heat capacity of the aluminium is

A. 200 JK^-1  B. 900 JK^-1  C. 90 JK^-1   D. 100 JK^-1

Answer

Heat capacity = quantity of energy supplied/temperature change = power x time (in seconds)/ θ

= 30 x 5 x 60 / 10 = 900 JK^-1

Question 9

A heating coil rated 1000W is used to boil off completely 2kg of boiling water. The time required to boil off the water is

A. 1.15 x 10⁴ s   B. 1.15 x 10³ s   C. 4.6 x 10⁴ s   D. 4.6 x 10³ s

[specific latent heat of vaporization of water = 2.3 x 10⁶ Jkg^-1]

Answer

Q = Pt = ML

1000 x t = 2 x 2.3 x 10⁶

T = 4.6 x 10³ = 4600 s

Question 10

An electric heater rated 220V, 1000W is immersed into a bucket full of water. Calculate the mass of water if the temperature changes from 30⁰C to 100⁰C and the current flows for 300 seconds.

A. 4.28kg   B. 42.86kg   C. 1.02kg   D. 7.14kg

[Specific heat capacity of water = 4200 J kg-1 K-1]

Answer

Pt = MCθ

1000 x 300 = M x 4200 x (100 – 30)

M = 300000 /(4200 x 70) = 1.02 kg

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Notes on conduction through liquid and gases

CONDUCTION THROUGH LIQUIDS

Electrolysis: Is the chemical change in a liquid due to the flow of electric current.

Electrolytes Are liquid that are good conductor of electricity.

Non-electrolytes are liquid that are poor conductor of electricity.

Application of Electrolysis

1. Electroplating: Is the process of coating metals with another metal in order to protect the metal from corrosion.
2. Purification of Metal: Electrolysis is important in purifying impure copper.

Faraday’s Laws of Electrolysis

Faraday’s First Law: States that the mass of a substance, M, of a substance liberated during electrolysis is strictly proportional to the quantity of electricity that has passed through the electrolyte.

Therefore:

M      =        Mass of a substance

Q       =        Quantity of electricity

M      =        zIt

Q = It

where z constant of proportionality which is the electrochemical equivalence

Faraday’s Second Law: States that he masses of the different substance deposited or liberated by the same quantity of electricity are strictly proportional to the equivalent of the substance.

Example 1

At what time must a current of 20A pass through a solution fo zinc sulphate to deposit 3g of zinc? If the electrochemical equivalent z = 0.0003387gc^-1.

I        =        20A

m = 3g

Z       =        0.0003387

M      =        zit

t        =        M/ ZI

t        =     3 / 0.0003387*20

t        =        443s

Gases

Conditions under which gases conduct electricity

1. Low pressure
2. High potential difference

Thermionic Emission

Thermionic emission: This is the emission of electrons from the surface of a hot metal. Whenever a metal is heated to a greater temperature, electrons are emitted from the surface of the metal in a process known as thermionic emission.

Application of thermionic emission

The application of thermionic emission is in the cathode –ray oscilloscope used for studying all types of waveforms, it is also important in measuring frequencies and amplitudes of voltages of electronic devices.

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