A student carried out an experiment to measure c, the specific heat capacity of water - Leaving Cert Physics - Question 4 - 2022

The student supplied heat energy using a Joule meter or by placing a heater below the calorimeter. The heat energy supplied can be quantified using the formula:

$ext{Heat energy (J)} = ext{Voltage (V)} imes ext{Current (I)} imes ext{Time (t)}$

The mass of the water can be calculated by subtracting the mass of the empty calorimeter from the total mass:

$ext{Mass of water} = ext{Mass of calorimeter and cold water} - ext{Mass of empty calorimeter}$

Substituting the values: $ext{Mass of water} = 0.1498 ext{ kg} - 0.0745 ext{ kg} = 0.0753 ext{ kg}$

The increase in temperature can be calculated as follows:

$ext{Increase in temperature} = ext{Final temperature} - ext{Initial temperature}$

Substituting the provided values: $ext{Increase in temperature} = 23 °C - 18 °C = 5 °C$

The formula used to calculate the heat gained by a material as it changes temperature is:

$Q = mc heta$

where:

• $Q$ is the heat gained (in joules),
• $m$ is the mass (in kg),
• $c$ is the specific heat capacity (in J kg⁻¹ K⁻¹),
• $heta$ is the change in temperature (in K or °C).

To find the specific heat capacity of water, we can rearrange the formula:

$c = \frac{Q}{m heta}$

Using the previously calculated values:

• $Q = 1703 ext{ J}$ (total heat energy added),
• $m = 0.0753 ext{ kg}$ (mass of water),
• $heta = 5 ext{ K}$ (increase in temperature).

Substituting these into the formula,

$c = \frac{1703 ext{ J}}{0.0753 ext{ kg} imes 5 ext{ K}} = \frac{1703}{0.3765} \approx 4525.75 ext{ J kg⁻¹ K⁻¹}$