Measurement of the specific latent heat of vaporisation of water

Physics Experiments

Apparatus

Calorimeter, lagging, beaker, conical flask fitted with stopper and delivery tube or steam generator, steam trap, retort stand, heat source, thermometer accurate to 0.1° C and electronic balance.

Procedure

• Half fill the conical flask or steam generator with water and fit with the delivery tube.

• Heat until steam issues freely.

• Find the mass of the calorimeter $m_{cal}$.

• Half fill the calorimeter with water cooled to approximately 10° C below room temperature.

• Find the mass $m_1$ of the water plus calorimeter.

• The mass of the cooled water $m_w$ is $m_1 - m_{cal}$.

• Record the temperature of the calorimeter plus water $θ_1$.

• Allow dry steam to pass into the water in the calorimeter until the temperature has risen by about 20° C.

• Remove the steam delivery tube from the water, taking care not to remove any water from the calorimeter in the process.

• Record the final temperature $θ_2$ of the calorimeter plus water plus condensed steam. The fall in temperature of the steam $Δθ_s$ is 100° C - $θ_2$.

• The rise in the temperature of the calorimeter plus water $Δθ$ is $θ_2 - θ_1$.

• Find the mass of the calorimeter plus water plus condensed steam m2. Hence the mass of the condensed steam $m_s$ is $m_2 - m_1$.

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Measurement of the specific latent heat of vaporisation of water

Results

Mass of the calorimeter $m_{cal} =$

Mass of the calorimeter plus the water $m_1 =$

Mass of the cooled water $m_w = m_2 - m_{cal} =$

Temperature of the calorimeter plus water $θ_1 =$

Final temperature of the calorimeter plus water $θ_2 =$ plus condensed steam

Fall in temperature of the steam $Δθ_1 = 100°C - θ_2 =$

Rise in the temperature of the calorimeter plus water $Δθ_2 = θ_2 - θ_1 =$

Mass of the calorimeter plus water plus condensed $m_2 =$ steam Mass of the condensed steam $m_s = m_2 - m_1 =$

Calculations

Assume heat losses to the surroundings cancel heat gains from the surroundings. Given that the specific heat capacity of water cw and the specific heat capacity of copper cc are already known, the specific latent heat of vaporisation of water l may be calculated from the following equation:

Energy lost by steam = energy gained by calorimeter + energy gained by the water

$m_s l + m_s c_w Δθ_1 = m_{cal} c_c Δθ_2 + m_w c_w Δθ_2$