In an experiment to measure the specific latent heat of vaporisation of water, cool water was placed in a polystyrene cup - Leaving Cert Physics - Question 2 - 2015

1. It is a good/neglectible heat insulator: This means that heat does not escape significantly from the cup.
2. It has a low/negligible specific heat capacity: This implies that the cup does not absorb a significant amount of heat during the experiment.

First, find the change in mass of the water:

• Initial mass of water, $m_{1} = 84.6 ext{g} - 1.2 ext{g} = 83.4 ext{g} = 0.0834 ext{kg}$

• Final mass of water, $m_{f} = 87.2 ext{g} - 1.2 ext{g} = 86.0 ext{g} = 0.0860 ext{kg}$

• Change in mass of water vaporised, $m = m_f - m_1 = 0.0860 ext{kg} - 0.0834 ext{kg} = 0.0026 ext{kg}$

Using the specific heat formula:

$Q = mc riangle T$

Where:

• $Q$ is energy gained
• $m$ is mass of water
• $c = 4180 ext{ J kg}^{-1} ext{K}^{-1}$
• Change in temperature, $riangle T = 30°C - 11°C = 19°C$

Calculating the energy gained by the water:

$Q = 0.0834 imes 4180 imes 19 = 6.82 imes 10^{3} ext{ J}$

Next, use the formula for latent heat:

L = rac{Q}{m} = rac{6.82 imes 10^{3}}{0.0026} = 2.62 imes 10^{6} ext{ J/kg}

Ensuring that only dry steam is used prevents any excess water from condensing and affecting the mass measurement. If moist steam were present, it would add additional mass to the system and potentially affect the temperature readings, leading to inaccuracies in calculating the latent heat.

Cooling the initial water ensures that all the heat added to the water comes from the steam alone. This avoids heat loss or gain from other sources, thereby providing more accurate data regarding the total heat exchanged during the experiment.