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 st...

Assessment: WAEC SSCE - Physics - 2016 (Practical) Subject: Physics

Question 1 Report

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 \(350cm^{3}\).

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)

Answer Details

Test of Practical Knowledge: heat carried into water by a hot mass

For each volume of water the mass is \(m=\rho v\) with \(\rho=1\ \text{g cm}^{-3}\), so numerically \(m/\text{g}=v/\text{cm}^3\). The temperature rise is \(\theta=\theta_1-\theta_0\), the initial temperature being \(\theta_0=34^{\circ}\text{C}\) in every trial.

(x) Table of readings

\(v\) / cm\(^3\)\(m=\rho v\) / g\(\theta_0\) / \(^{\circ}\)C\(\theta_1\) / \(^{\circ}\)C\(\theta=\theta_1-\theta_0\) / \(^{\circ}\)C
150.0150.0348854
200.0200.0348652
250.0250.0348450
300.0300.0348046
350.0350.0347642

(xi) Graph of \(m\) against \(\theta\)

Plotting \(m\) on the vertical axis against \(\theta\) on the horizontal axis gives a straight line of best fit with a negative gradient: the larger the mass of water, the smaller the temperature rise produced by the same 50 g hot mass.

graph
Graph of m (vertical axis) against θ (horizontal axis); the line of best fit has a negative gradient of magnitude 16.2 g °C⁻¹.

(xii) Slope of the graph

Two points read off the line of best fit are \((\theta=42^{\circ}\text{C},\,m=360\ \text{g})\) and \((\theta=54^{\circ}\text{C},\,m=166\ \text{g})\).

\[|s|=\left|\frac{\Delta m}{\Delta\theta}\right|=\frac{360-166}{54-42}=\frac{194}{12}=16.2\ \text{g }^{\circ}\text{C}^{-1}.\]

The gradient is negative, \(s=-16.2\ \text{g }^{\circ}\text{C}^{-1}\); its magnitude is used below.

(xiii) Evaluation of \(k\)

\[k=\frac{50}{|s|}=\frac{50}{16.2}=3.1.\]

(xiv) Two precautions

  • I stirred the water gently and avoided splashing so that heat was evenly distributed before the highest temperature was read.
  • I avoided parallax error by reading the thermometer and the measuring cylinder with the eye level with the meniscus, and I transferred the hot mass quickly and gently into the water.

(b)(i) Heat capacity

The heat capacity of a body is the quantity of heat required to raise the temperature of the whole body by one kelvin (one degree Celsius). Its SI unit is the joule per kelvin, \(\text{J K}^{-1}\).

(b)(ii) Time to reach boiling point

Data: \(P=1.2\ \text{kW}=1200\ \text{W}\), \(m=800\ \text{g}=0.8\ \text{kg}\), \(c=4200\ \text{J kg}^{-1}\text{K}^{-1}\), temperature rise \(\Delta\theta=101-20=81\ \text{K}\).

Neglecting heat losses, the electrical energy supplied equals the heat gained by the water:

\[Pt=mc\,\Delta\theta\] \[t=\frac{mc\,\Delta\theta}{P}=\frac{0.8\times4200\times81}{1200}=\frac{272160}{1200}=226.8\ \text{s}.\]

The kettle takes about 226.8 s (approximately 3.8 minutes) to bring the water to its boiling point.

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