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

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.

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

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

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

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

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

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

From the data:

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

This gives:

$R = \frac{6.50}{0.09 \times 4180} \approx 2.06 \Omega$