WAEC Physics Past Questions & Answers - Page 229

1,141.

A cup containing 100g of pure water at 20°C is placed in a refrigerator. If the refrigerator extracts heat at a rate of 840J per minute, calculate the time taken for the water to freeze. Neglect the heat capacity of the material of the cup. (Specific heat capacity of water =4.2J\(g^{-1}K^{-2}\))((Specific latent heat fusion of water =336J\(g^{-1}\)

15 minutes

20 minutes

42 minutes

50 minutes

84 minutes

View answer & explanation

Correct answer is D

H=pt =mL +mcθ ; 840t= 100x336+100x4.2x20 thus t=420000/840=50minutes

1,142.

A converging lens of focal length 15cm is used to obtain a real image magnified \(1\frac{1}{2}\) times. Calculate the distance of the image from the lens?

37.5cm

22.5cm

15cm

7.5cm

3.3cm

View answer & explanation

Correct answer is A

If u=x, v = \(1\frac{1}{2}x \Rightarrow if u= 2x,v=3x\\
\frac{1}{f}=\frac{1}{u}+\frac{1}{v}; \frac{1}{15}=\frac{1}{2x}+\frac{1}{3x} \times \frac{5}{6}\\
X=\frac{75}{76}=12.5, v=3x=12.5\times 3=37.5\)

1,143.

A fixed mass of gas occupies a volume of 20\(cm^3\) at a fixed pressure of 700mmHg. Assuming that the temperature remains constant, what will be the volume of the gas at 750mmHg?

2.5\(cm^3\)

15.5\(cm^3\)

18.7\(cm^3\)

21.4\(cm^3\)

72.5\(cm^3\)

View answer & explanation

Correct answer is C

\(v_2=\frac{v_1p_1}{p_2}=\frac{20\times700}{750}=18.66=18.7cm^3\)

1,144.

Calculate the quantity of heat released when 100g of steam at 100°C condenses to water(Take the specific latent heat of vaporization of water as 2.3x \(10^6Jkg^{-1}\))

2.3x\(10^{1}J\)

2.3x\(10^{2}J\)

2.3x\(10^{4}J\)

2.3x\(10^{5}J\)

3.3x\(10^{7}J\)

View answer & explanation

Correct answer is D

\(\Delta{H}=mL=0.1\times2.3\times10^6=2.3\times10^5\)J

1,145.

A solid material of volume 100\(cm^{-3}\) is heated through a temperature difference of 40°C. Calculate the increase in the volume of the material if its linear expansitivity is 2 x \(10^{-6}K{-1}\)

2.4 x\(10^{-2}cm^{-3}\)

1.6 x\(10^{-2}cm^{-3}\)

8 x\(10^{-2}cm^{-3}\)

5 x\(10^{-2}cm^{-3}\)

4 x\(10^{-2}cm^{-3}\)

View answer & explanation

Correct answer is A

\(3{\alpha}= \frac{\Delta{V}}{V_1\Delta{T}}\Rightarrow \frac{\Delta{V}}{100\times40}=3\alpha\, where\, \alpha = 2 \times 10^{-6}\\ \frac{\Delta{V}}{4000}=6\times10^{-6}\Rightarrow \Delta=4\times10^3\times6\times10^{-6}= 2.4\times10^{-2}cm^{-3}\)