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In an experiment to verify Joule's law, a constant current, I, was passed through a heating coil immersed in water - Leaving Cert Physics - Question 4 - 2018

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In an experiment to verify Joule's law, a constant current, I, was passed through a heating coil immersed in water. The current was allowed to flow for 3 minutes and... show full transcript

Worked Solution & Example Answer:In an experiment to verify Joule's law, a constant current, I, was passed through a heating coil immersed in water - Leaving Cert Physics - Question 4 - 2018

Step 1

Draw a diagram of the apparatus used in this experiment.

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Answer

To illustrate the apparatus for the experiment verifying Joule's law, draw the following components:

  • A power supply (represented as a battery).
  • A variable resistor (to adjust the current).
  • An ammeter (to measure the current in series).
  • A heating coil immersed in a beaker or container with water.
  • A thermometer to monitor the water temperature.

Label each component clearly for clarity and completeness.

Step 2

Draw a suitable graph to verify Joule's law.

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To construct a graph for verifying Joule's law, plot the following:

  • Labelled axes:

    • X-axis: Current squared (I²) values with appropriate scaling.
    • Y-axis: Change in temperature (Δθ) in °C.
  • Points plotted:

    • Calculate I² for each I value:
      • For I = 1.5 A, I² = 2.25
      • For I = 2.0 A, I² = 4.00
      • For I = 2.5 A, I² = 6.25
      • For I = 3.0 A, I² = 9.00
      • For I = 3.5 A, I² = 12.25
      • For I = 4.5 A, I² = 20.25
      • For I = 5.5 A, I² = 30.25
    • Corresponding Δθ values recorded from the table.
  • Use a straight line with good fit through the origin to connect the points plotted.

Step 3

Calculate the slope of the graph and hence calculate the resistance of the heating coil.

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To find the slope of the graph, use the formula:

slope=y2y1x2x1\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}

Selecting points from the graph (for example, (2.0, 4.0) and (4.5, 20.25)):

slope=20.254.04.52.0=16.252.5=6.50\text{slope} = \frac{20.25 - 4.0}{4.5 - 2.0} = \frac{16.25}{2.5} = 6.50

Using the slope to calculate the resistance (R) of the heating coil, and applying Joule’s law:

R=RTmcR = \frac{R T}{m c} where R = slope,

From the data:

  • R=6.50R = 6.50
  • m=0.09kgm = 0.09 kg (90 g), c=4180J/kg/Kc = 4180 J/kg/K

This gives:

R=6.500.09×41802.06ΩR = \frac{6.50}{0.09 \times 4180} \approx 2.06 \Omega

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