Day: December 6, 2022

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202212061213 Solution to 1976-HL-PHY-I-4

(a) A kettle of negligible heat capacity, full of liquid, was placed over a burner. It was found that the liquid, initially at a temperature of 297\,\mathrm{K}, reached the boiling point of 378\,\mathrm{K} in 9 minutes, and after another 60 minutes all the liquid in the kettle boiled away. Neglecting heat loss, calculate the latent heat of vaporisation of the liquid. Specific heat capacity of the liquid =\textrm{4,000}\,\mathrm{J\,kg^{-1}\,K^{-1}}.

(b) Describe another method to determine the latent heat of vaporisation of a liquid.

Roughwork.

(a)

Let P be the power of the burner, and m the mass of the liquid. Assume no heat loss to the surroundings, then

\begin{aligned} Pt = E & = Q = mc\Delta T \\ P\times 9(60) & = m\times 4000\times (378-297) \\ \frac{P}{m} & = 600 \\ \end{aligned}

Suppose for the liquid the latent heat of vaporisation is l_v, then

\begin{aligned} Pt = E & = Q = ml_v \\ P\times 60(60) & = ml_v \\ l_v & = 3600\times\bigg(\frac{P}{m}\bigg) \\ & = 3600\times (600) \\ & = 2.16\times 10^6\,\mathrm{J\,kg^{-1}} \\ \end{aligned}

(b) Left as an exercise to the reader.