A student performed a single experiment to (a) verify Joule’s law, and (b) determine the specific heat capacity of olive oil - Leaving Cert Physics - Question 5 - 2022

The mass of the olive oil was determined by:

1. Using a balance to weigh an empty calorimeter.
2. Filling the calorimeter with olive oil until it reached the desired volume.
3. Weighing the filled calorimeter to find the total mass.
4. Subtracting the mass of the empty calorimeter from the total mass to obtain the mass of the olive oil.

To graphically verify Joule's law:

1. Plot your data: Use the calculated power (P = I^2 * R) on the y-axis and the temperature change (Δθ) on the x-axis using the given current values.
2. Label the axes: Ensure the y-axis is labeled 'Power (W)' and the x-axis is labeled 'Temperature Change (°C)'.
3. Include a line of best fit: Draw a straight line that best represents the data points, confirming that the relationship between power and temperature change is linear.

To calculate the slope (m) of the graph:

1. Choose two points on the line of best fit, (x1, y1) and (x2, y2).
2. Use the formula for slope: $m = \frac{y_2 - y_1}{x_2 - x_1}$
3. Substitute in the values from the line of best fit to find the slope.

To calculate the specific heat capacity (c) of olive oil:

1. Use the formula derived from Joule’s law: $Q = mc\Delta\theta$ where Q is the heat transferred, m is the mass, c is the specific heat capacity, and Δθ is the temperature change.

2. From the graph, use the slope calculated as the increase in power per degree change in temperature. Rearranging the formula, we find: $c = \frac{Q}{m \Delta \theta}$ With the previously determined values for Q and Δθ, substitute into the equation.

3. Carefully substitute the known quantities to compute the specific heat capacity: $c = \frac{0.447 \times 8.5 \times 180}{0.35} = 1954 \text{ J kg}^{-1} ext{ K}^{-1}$