A student carried out an experiment to measure l, the specific latent heat of vaporisation of water - Leaving Cert Physics - Question 4 - 2023

To find the mass of cold water (A): $A = m_{cal + water} - m_{cal} = 0.1327 ext{ kg} - 0.0894 ext{ kg} = 0.0433 ext{ kg}$

For mass of added steam (B): $B = m_{final} - m_{initial} = 0.1341 ext{ kg} - 0.1327 ext{ kg} = 0.0014 ext{ kg}$

To calculate the increase in temperature (C): $C = T_{final} - T_{initial} = 36 °C - 20 °C = 16 °C$

To find the decrease in temperature (D) of the steam: $D = 100 °C - T_{final} = 100 °C - 36 °C = 64 °C$

According to the energy balance: Heat lost by steam = Heat gained by water and calorimeter: $B imes l + (0.1327 - 0.0894) imes 4200 imes 16 = 0.0894 imes 390 imes 64$ Substituting the values: $0.0014 imes l + 0.0433 imes 4200 imes 16 = 0.0894 imes 390 imes 64$ Solving the equation will give: $l = ext{value after calculating}$