Physic Tutorials

How to solve questions on Heat energy for UTME

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

Clik to to read on Heat energy

inspir

Bolarinwa Olajire

A lecturer, Educationist, PhD student at FUNAAB, and a Blogger.

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