JAMB Physics Past Questions & Answers - Page 29

141.

Calculate the heat energy required to vaporize 50g of water initially at 80ºC if the specific heat capacity of water is 4.23jg\(^{-1}\)k\(^{-1}\) (specific latent heat of vaporization of water is 2260jg\(^{-1}\)

A.

530000J

B.

23200J

C.

17200J

D.

130000J

Correct answer is D

H = mc\(\bigtriangleup\theta\) + mL

H = ?

m = 50g, c = 4.23J/gk,L = 2260, \(\bigtriangleup\theta\) = 80ºC

H = 50 [4.23 x 80 + 2260]

= 129920 \(\approx\) 130,000J

142.

A boy standing some distance from the foot of a tall cliff claps his hands and hears an echo 0.5s later. If the speed of sound is 340ms\(^{-1}\), how far is he from the cliff?

A.

680m

B.

170m

C.

34m

D.

85m

Correct answer is D

V = \(\frac{2d}{t}\) 
v = 340ms\(^{-1}\)
t = 0.5s
d= ?

d = \(\frac{2d}{2} = \frac{340 \times 0.5}{2}\) = 85m

143.

A source of sound produces waves in air of wavelength 1.65m. If the speed of sound in air is 330ms\(^{-1}\), the period of vibration in air is?

A.

200

B.

0.005

C.

0.5

D.

0.02

Correct answer is B

V = \(\frac{\lambda}{T}\)

: v = 330ms\(^{-1}\)
 \(\lambda\) = 1.65m
 T = ?

330 = 1.65

T = \(\frac{1.65}{330}\) = 0.005s

144.

Which of the following instruments may be used to measure relative humidity?

A.

Hydrometer

B.

Manometer

C.

Hygrometer

D.

Hypsometer

Correct answer is C

A hygrometer is an instrument that is used to measure water vapor in the atmosphere. This is also commonly referred to as relative humidity in the atmosphere.

145.

0.5kg of water at 10ºC is completely converted to ice at 0ºC by extracting (88000) of heat from it. If the specific heat capacity of water is 4200jkg\(^{1}\) Cº. Calculate the specific latent heat of fusion of ice.

A.

9.0kjkg\(^{-1}\)

B.

84.0kjkg\(^{-1}\)

C.

134.0kjkg\(^{-1}\)

D.

168.0kjkg\(^{-1}\)

Correct answer is C

Heat involved (H) = 88,000J, Mass(M)= 0.5kg,  specific heat capacity of water(C) = 4200JkgºC,

Specific latent heat of fusion(L) = ?, Temperature change(Δθ) =  θ2 -  θ1 =  (10 - 0)° = 10°

H = MCΔθ + ML 

or

H = M(CΔθ + L) -->\(\frac{H}{M}\) = CΔθ + L 

: L = \(\frac{H}{M}\) -  CΔθ = \(\frac{88,000}{0.5}\) - 4200 \(\times\) 10

L = 176,000 - 42000 = 134,000

L = 134,000 or 134kj/kg