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. The polystyrene cup is a good insulator, meaning it does not transfer heat effectively to the surroundings.
2. The cup has a negligible heat capacity compared to the heat absorbed by the water, implying it does not gain heat.

1. Calculate the mass of water used:

   $m_w = m_{initial} - m_{final} = 84.6 ext{ g} - 87.2 ext{ g} = 0.0068 ext{ kg}$

2. The temperature change of water: $\Delta T = T_{final} - T_{initial} = 30 °C - 11 °C = 19 °C$

3. The energy gained by the water:

   $E = m_w imes c imes \Delta T = 0.0068 ext{ kg} imes 4180 ext{ J kg}^{-1} ext{ K}^{-1} imes 19 ext{ K} = 0.25 imes 10^4 ext{ J}$

4. Finally, use the formula for specific latent heat:

   $L = \frac{E}{m_{steam}}$

Assuming all steam is condensed, where we calculate the specific latent heat.

Dried steam ensures that only steam is added, removing the possibility of introducing any additional water mass, which would otherwise alter the heat calculations. Cooling the initial water prevents any additional heat absorption or loss, ensuring that variations in temperature are minimized.

By ensuring that the steam added was entirely dry and that the initial water was cooled, the student eliminated variables that could skew the heat calculations. This allows for more accurate measurements of energy transfer and thus a more reliable calculation of the specific latent heat of vaporization.