Question

A burner supplies heat energy at a rate of $434{\mathrm{Js}}^{-1}$ for $60$ seconds when $40\mathrm{g}$ of ice at $0°\mathrm{C}$ changes to water at $75°\mathrm{C}$. Calculate the latent heat of ice.

Answer 1

Step 1: Given data

Heat energy supplied by the burner$=434J{s}^{-1}$for an interval of $60s$.

Mass of ice$=40g$

Step 2: Finding the heat energy supplied by the burner

Heat energy is $Q=P\times t$.

Substituting the values,

$Q=434\times 60=26040J$

Step 3: Finding the latent heat of ice ($L_i$)

Heat energy furnished by the burner $=$ Heat used in melting the ice and converting into water at $75°\mathrm{C}$.

The heat used for melting ice and converting into water will be equal to the sum of heat energy for melting ice and the heat energy for converting it into water.

This total heat energy will be then equal to the heat energy produced by the burner.

$Q=m{\mathrm{L}}_{\mathrm{i}}+m{\mathrm{c}}_{\mathrm{w}}(75-0)$

Specific heat capacity of water is $c_w=4.2J/g°C$, so putting the values,

$$\begin{gathered} 26040=40L_i+40\times 4.2\times 75 \\ 40{\mathrm{L}}_{\mathrm{i}}=26040-12600 \\ 40{\mathrm{L}}_{\mathrm{i}}=13440 \\ {L}_i=\frac{13440}{40}=336J{g}^{-1} \end{gathered}$$

Hence, the latent heat of ice is $336J{g}^{-1}$.