91Ó°ÊÓ

1. Mass of water,${\mathrm{M}}_{\mathrm{W}}=0.530\mathrm{kg}$
2. Specific heat of water,${\mathrm{C}}_{\mathrm{W}}=4190\mathrm{J}/{\mathrm{kg}}^{\mathrm{o}}\mathrm{C}$
3. Change in temperature,$∆\mathrm{T}={40}^{\mathrm{o}}\mathrm{C}$
4. Specific heat of the ice,${\mathrm{C}}_{\mathrm{i}}=2220\mathrm{J}/{\mathrm{kg}}^{\mathrm{o}}\mathrm{C}$

A substance's specific heat capacity is the amount of energy required to raise its temperature by one degree Celsius. Using the formula for specific heat, we can find the amount of heat, and using the formula for the power in terms of energy and time, we can find the power. Also, using the condition for the equilibrium, we can find the initial and final mass of the ice.

Formula:

The heat energy required by a body,$\mathrm{Q}=\mathrm{mc}∆\mathrm{T}$ …(¾±)

Where, m = mass

c = specific heat capacity

$∆\mathrm{T}$= change in temperature

Q= required heat energy

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

Where, m = mass

L = specific latent heat

The rate of the heat transferred by the body, $\mathrm{P}=\frac{\mathrm{Q}}{\mathrm{t}}$ …(¾±¾±¾±)

The heat generated by the body can be calculated by using equation (i) as:

$\begin{array}{rcl}\mathrm{Q}& =& 0.530\mathrm{kg}×4190\mathrm{J}/{\mathrm{kg}}^{\mathrm{o}}\mathrm{C}×{40}^{\mathrm{o}}\mathrm{C}\\ & =& 88828\mathrm{J}\end{array}$

The rate P can be calculated by using equation (iii) as:

$\begin{array}{rcl}\mathrm{P}& =& \frac{88828\mathrm{J}}{40\min }\\ & =& \frac{88828\mathrm{J}}{2400\mathrm{s}}\\ & =& 37\mathrm{W}\end{array}$

Hence, the rate P is 37 W

During the same time of 40 mins the ice warms by${20}^{\mathrm{o}}\mathrm{C}$

We have the formula of equation (i), and the mass of the ice is given as:

$\begin{array}{rcl}{\mathrm{M}}_{\mathrm{ice}}& =& \frac{\mathrm{Q}}{{\mathrm{C}}_{\mathrm{ice}}∆\mathrm{T}}\\ & =& \frac{88828\mathrm{J}}{2220\mathrm{J}/{\mathrm{kg}}^{\mathrm{o}}\mathrm{C}×{20}^{\mathrm{o}}\mathrm{C}}\\ & =& 2.0\mathrm{kg}\end{array}$

Hence, the initial mass of the ice in the container is 2.0 kg

To find the ice produced by freezing the water that has already$0^{\mathrm{o}}\mathrm{C}$temperature, we have to use the time span between 40 min to 60 min

The heat produced during the process of 20 min is given as:

$\begin{array}{rcl}{\mathrm{Q}}_{20\min }& =& \frac{{\mathrm{Q}}_{40\min }}{2}\\ & =& \frac{88828\mathrm{J}}{2}\\ & =& 44414\mathrm{J}\end{array}$

Now put the value of Q in the formula (ii), we get the mass of the ice produced as:

$\begin{array}{rcl}{\mathrm{M}}_{\mathrm{water}\mathrm{becoming}\mathrm{ice}}& =& \frac{{\mathrm{Q}}_{20\min }}{{\mathrm{L}}_{\mathrm{f}}}\\ & =& \frac{44414\mathrm{J}}{333×{10}^3\mathrm{J}/\mathrm{kg}}\\ & =& 0.13\mathrm{kg}\end{array}$

Hence, the mass of the ice is 0.13 kg