Question

How much water remains unfrozen after$\mathbf{50}\mathbf{.2}\mathbf{}\mathbf{\text{kJ}}$is transferred as heat from$\mathbf{260}\mathbf{}\mathbf{\text{g}}$of liquid water initially at its freezing point?

Short answer

The amount of water that still remains frozen is$109\times {10}^{-3}\text{kg}$

Step-by-step solution

The given data

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

Understanding the concept of heat transformation

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, $Q=Lm$ …(i)

Where,m = mass

L = specific latent heat

Calculation of the amount of unfrozen water

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

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

So the mass of the frozen water is given as:

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

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

$260\times {10}^{-3}\text{kg}-151\times {10}^{-3}\text{kg}=109\times {10}^{-3}\text{kg}$

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