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2016 WAEC Physics Practical You are provided with a beaker, a thermometer, a stirrer, a measuring cylinder, a bunsen...

Physics
WAEC 2016

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You are provided with a beaker, a thermometer, a stirrer, a measuring cylinder, a bunsen burner, a wire gauze, a 50g mass, a pair of tongs, water, tripod stand, and other necessary materials.

i. Using the measuring cylinder, measure 150cm\(^{3}\) of water into the beaker.

ii. Record the volume v of the water in the beaker

iii. Calculate the mass m of the water, given that m = pv and; p = 1gcm\(_{-3}\). 

iv. Measure and record the initial temperature \(\theta_{0}\) of the water in the beaker.

v. Hold the 50g mass with the pair of tongs in the flame of the bunsen burner for 2 minutes.

vi. Quickly transfer the 50g mass to water in the beaker.

vii. Stir gently and record the highest temperature \(\theta_{1}\), attained

viii. Evaluate \(\theta\) = (\(\theta_{1}\) - \(\theta_{0}\)).

ix. Empty the content of the beaker and repeat the procedures above for the values of v = 200cm\(^{3}\), 250cm\(^{3}\), 300cm\(^{3}\), and 350 cm\(^{3}\).

x. Tabulate your readings.

xi. Plot a graph with m on the vertical axis and \(\theta\) on the horizontal axis.

xii. Determine the slope, s, of the graph.

xiii. Evaluate k = \(\frac{50}{s}\).

xiv. State two precautions taken to obtain accurate results.

(b)i. Define heat capacity.

ii. An electric kettle rated 1.2kw is used to heat 800g of water initially at a temperature of 20 C. Neglecting heat losses, calculate the time taken for the kettle to heat the water to its boiling point. [Take the boiling point of water= 101 C specific heat capacity of water = 4200 Jkg' K'1 (odv)

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Explanation
S/N m/g T/min \(\theta\) = \(\theta_{1} - \theta_{0}\)
1 150.0 2.0 \(\theta\) = 88 - 34 = 56\(^{6}\)
2 200.0 4.0 \(\theta\) = 86 - 34 = 52
3 250.0 6.0 \(\theta\) = 84 - 34 = 50
4 300.0 8.0 \(\theta\) = 80 - 34 = 46
5 350.0 10.0 \(\theta\) = 76 - 34 = 42

Initial temperature of water = 34°

Slope s = \(\frac{\bigtriangleup {m}}{\bigtriangleup \theta}\)

    = \(\frac{m_{2} - m_{1}}{\theta_{2} -\theta_{1}}\)

    = \(\frac{350 - 50}{62 - 42} = \frac{300}{20}\)

S = 150

Evaluate k = \(\frac{50}{S} = \frac{50}{150}\) = 0.333

Precautions;
- Avoided splashing of water
- Stirred continuously (for even distribution of heat).
- Avoided parallax error on the thermometer/measuring cylinder
- Repeated readings shown on the table
- Ensured the heated mass was gently dropped into the beaker.

(b)i. Heat Capacity Is the quantity of heat required to raise the temperature of a substance.

ii. Pt = mc\(\theta\)

1200 x t = \(\frac{800}{1000}\) x 4200 (101 x 20)

1200t = 0.8 x 4200 x 81

t = \(\frac{0.8 \times 4200 \times 81}{1200}\)

t = \(\frac{272160}{1200}\)

t = 2260.8 secs

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