Question

$500g$ of molten lead at its melting point is just kept in a molten state by a heater of $25W$. If the heater is switched off, the temperature of lead starts falling after $7min$. Calculate the specific latent heat of fusion of lead.

Answer 1

Step 1: Given data

1. The quantity of molten lead is $500g$
2. The power of the heater is $25W$

Step 2: Calculating the heat given out by lead

Let the specific latent heat of lead is $LJ{g}^{-1}$.

Heat given out by lead at its melting point

$$\begin{gathered} =amountoflead\times Specificlatentheatoflead \\ =500\times L \end{gathered}$$

The heat is given by the lead

$$\begin{gathered} =(Powerofheater\times time) \\ =(25\times 7\times 60)J \\ =10500J \end{gathered}$$

Step 3: Calculating the specific latent heat of fusion of lead

By the principle of calorimetry

$$\begin{gathered} \text{Heat given out by lead at its melting point = Heat given out by lead} \\ 500\times L=10500 \\ \Rightarrow L=\frac{10500}{500} \\ \Rightarrow L=21J{g}^{-1} \end{gathered}$$

Hence, the latent heat of lead is $21J{g}^{-1}$.