TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with a beaker, a thermometer, a stirrer, a measuring cylinder, a bunsen burner, a wire gauze, a 50g ma...
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)
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
150.0
150.0
34
88
54
200.0
200.0
34
86
52
250.0
250.0
34
84
50
300.0
300.0
34
80
46
350.0
350.0
34
76
42
(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 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 \((\theta=54^{\circ}\text{C},\,m=166\ \text{g})\).
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, \(\text{J K}^{-1}\).
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
150.0
150.0
34
88
54
200.0
200.0
34
86
52
250.0
250.0
34
84
50
300.0
300.0
34
80
46
350.0
350.0
34
76
42
(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 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 \((\theta=54^{\circ}\text{C},\,m=166\ \text{g})\).
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, \(\text{J K}^{-1}\).
