Question: MaterialsA,B, andCare solids that are at their melting temperatures. MaterialArequires 200 Jto melt 4Kg, materialBrequires 300 Jto melt 5kg, and materialCrequires 300 Jto melt 6kg. Rank the materials according to their heats of fusion, greatest first.

• i) The material A requires heat Q= 200J to melt m = 4kg.
• ii) The material B requires heat Q=300J to melt m = 5kg .
• iii) The material C requires heat Q=300 J to melt m =6kg .

The amount of heat required to transform 1 g of a solid into a liquid without a temperature change is known as the heat of fusion. Using equation 18-16b for the total energy transformed, we can find the heat of fusion for each material and by comparing the respective values of the heat of fusion, and we can rank the materials according to their heat of fusion.

Formula:

The total energy transferred to the body,$Q=L_{F}m$ …(i)

where, $L_F$is the heat of fusion

Let,$L_{F,A},L_{F,B}and{L}_{F,C}$be the heats of fusion of material A, material B, and material C respectively.

The heat of fusion for material A is calculated using the given values in equation (i) as follows:

$L_{F,A}=\frac{200J}{4kg}$

Similarly,the heat of fusion for material B is calculated using the given values in equation (i) as follows:
$$\begin{gathered} L_{F,B}=\frac{300J}{5kg} \\ =60J/kg \end{gathered}$$

And the heat of fusion for material C is calculated using the given values in equation (i) as follows:
$$\begin{gathered} L_{F,C}=\frac{300J}{6kg} \\ =50J/kg \end{gathered}$$

Therefore, the ranking of the materials according to their latent heat of fusion is$L_{F,B}>L_{F,A}=L_{F,C}$.