91Ó°ÊÓ

The specific heat and the latent of the vaporisation are given per mass.

Knowing the eaquation

$m=\mathrm{ÒÏ}V$, the quantities of iron and water have to be converted to mass from volume and $\mathrm{ÒÏ}$is the mass densty in the equation.

Known factors:

The density of water, ${\mathrm{ÒÏ}}_w=1000\raisebox{1ex}{$Kg$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.$, and

the density of Iron, ${\mathrm{ÒÏ}}_i=7874\raisebox{1ex}{$Kg$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.$.

Given informations, role="math" localid="1649073521148" $$\begin{gathered} m_w=1000·200·1·{10}^{-6}=0.2kg \\ {m}_i=7874·65·1·{10}^{-6}=0.51181kg \end{gathered}$$

This experiment is done because one of the following three will happen after the iron ot water,

(1) The boiling point of the water will be greater than the Equilibrium temperature, in which no water will have boiled.

(2) The boiling point of the water and the equilibrium temperature will be equal.

(3) The boiling point of the water will be less than the equilibrium point of the water.

Let's assume that the 2nd case will happen. The entire water will be at the boiling point ny taking some energy from the hot iron. The rest energy from the hot iron will evaporate X part of the total mass($m_w$) of the water.

role="math" localid="1649074601941" $$\begin{gathered} X{m}_w{L}_w+c_w{m}_wâ–³T_w=c_i{m}_iâ–³T_i \\ X=\frac{c_i{m}_iâ–³T_i-c_w{m}_wâ–³T_w}{m_w{L}_w} \end{gathered}$$

Putting all the numerical values to get the value of X,

$X=\frac{449·0.51181·(800-100)-4190·0.2·(100-20)}{0.2·2.256·{10}^6}=0.2079$

Approximating and taking the ratio in percentage, the final answer will be,

$X=21\%$