JAMB Physics Past Questions & Answers - Page 465

2,321.

Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss;
Specific heat capacity of copper = 400JKg-1K-1
Latent heat of fusion of ice = 3.3 x 105JKg-1

A.

8/33kg

B.

33/80kg

C.

80/33kg

D.

33/8kg

Correct answer is A

Heat given off by the copper = MCt
= 2 x 400 x (100 - 0)
Heat absorbed by the ice = ML
= M x 3.3 x 105
Heat lost = Heat gained
∴ M x 3.3 x 105 = 2 x 400 x 100
∴ M = (2 x 400 x 100)/3.3 x 105
= 8/33Kg

2,322.

The temperature gradient across a copper rod of thickness 0.02m, maintained at two temperature junctions of 20°C and 80°C respectively is

A.

3.0 x 102Km-1

B.

3.0 x 103Km-1

C.

5.0 x 103Km-1

D.

3.0 x 104Km-1

Correct answer is B

Temperature Gradient = \(\frac{\Delta {\text{temp.}}}{\text{thickness}}\)

= \(\frac{80 - 20}{0.02}\)

= \(\frac{60}{0.02}\)

= 3000Km-1

= 3.0 x 103Km-1

2,324.

A piece of substance of specific head capacity 450JKg-1K-1 falls through a vertical distance of 20m from rest. Calculate the rise in temperature of the substance on hitting the ground when all its energies are converted into heat. [g = 10ms-2]

A.

2/9°C

B.

4/9°C

C.

9/4°C

D.

9/2°C

Correct answer is B

Energy at the height = mgh
This energy is converted to heat energy - mcΔt
∴ From the principle of energy conversion,
mcΔt = mgh
m x 450 x Δt = m x 10 x 20
∴ Δt = m x 10 x 20/(m x 450)
= 200/450
= 4/9°C

2,325.

A gas at a volume of V0 in a container at pressure P0 is compressed to one-fifth of its volume. What will be its pressure if the magnitude of its original temperature T is constant?

A.

P0/5

B.

4P0/5

C.

P0

D.

5P0

Correct answer is D

At constant T, P1 V1 = P2 V2.

Let P1 = P0 and V1 = V0, then solve accordingly.

As pressure is varies inversely as volume according to:

Boyle's law, also referred to as the Boyle–Mariotte law, or Mariotte's law, is an experimental gas law that describes how the pressure of a gas tends to decrease as the volume of the container increases.

As volume is reduced by one-fifth; pressure will increase by five times.