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# How to solve questions on Heat energy for UTME

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

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

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