91Ó°ÊÓ

1. The heat transferred from water is$Q=50.2\text{ }\text{kJor}50.2×{10}^3\text{J}$
2. The mass of the water is$m=260\text{gor}260×{10}^{−3}\text{kg}$

The theory of heat transfer aims to foretell the potential energy transfer between material bodies due to temperature differences. We can use the concept of the heat of transformation. The phase changes from solid to liquid, hence, the heat of transformation involved in this change, is called the heat of fusion.

Formula:

The heat energy released or absorbed by the body, $Q=Lm$ …(¾±)

Where,m = mass

L = specific latent heat

The expression of the heat of fusion water using equation (i) is given.

The heat of fusion of water,$L_F=333×{10}^3\text{J/kg}$

So the mass of the frozen water is given as:

$\begin{array}{c}m=\frac{Q}{L_F}\\ =\frac{50.2×{10}^3\text{J}}{333×{10}^3\text{J/kg}}\\ =0.151\mathrm{kg}\end{array}$

So, the mass of the unfrozen water is given as:

$260×{10}^{−3}\text{kg}−151×{10}^{−3}\text{kg}=109×{10}^{−3}\text{kg}$

Hence, the amount of unfrozen water is$109×{10}^{−3}\text{kg}$