91Ó°ÊÓ

i) Boiling point of ethyl alcohol is$78.0{\text{ }}^{\text{0}}\text{C}.$

ii) Freezing point of ethyl alcohol is$−114{\text{ }}^{\text{0}}\text{C}$

iii) Latent heat of vaporization of ethyl alcohol $\left(L_v\right)$is$879\text{ k´³/°ì²µ}$

iv) Mass of ethyl alcohol is$0.510\text{ k²µ}$

v) Heat of fusion is$109\text{ k´³/°ì²µ}$

vi) Specific heat Is $2.43\text{ k´³/°ì²µK}$

A substance's specific heat capacity is the amount of energy required to raise its temperature by one degree Celsius. The values of latent heat of vaporization, freezing point, boiling point, and also the mass of the ethyl alcohol are given. Using these values and the equation for specific heat capacity and latent heat, we can calculate the amount of heat removed from ethyl alcohol by using the equations.

Formula:

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

Where,$m$ = mass

$c$= specific heat capacity

$\mathrm{Δ}T$= change in temperature

$Q$= required heat energy

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

Where,

$L$= specific latent heat of vaporization

The phase change occurs at${78}^{\text{0}}\text{C}.$To find the heat required for this phase change, we can use the formula (ii) as given:

$\begin{array}{c}{Q}_1=\frac{879\text{ k´³}}{\text{kg}}×0.510\text{ k²µ}\\ =448.29\text{ k´³}\end{array}$

To cool the liquid to$−{114}^{\text{0}}\text{C}$the heat energy required for this phase can be given using the formula of equation (i) as given:

$\begin{array}{c}{Q}_2=2.43\frac{\text{kJ}}{\text{kg.K}}×0.510\text{ k²µ}×192\text{‿é}\\ =237.95\text{kJ}\end{array}$

Finally, to accomplish the phase change at,$−{114}^{\text{0}}\text{C}$ we use the formula of equation (ii), which is given as:

$\begin{array}{c}{Q}_3=\frac{109\text{kJ}}{\text{kg}}×0.510\text{ k²µ}\\ =55.59\text{kJ}\end{array}$

Therefore, the total amount of heat to be removed from ethyl alcohol is given as:

$\begin{array}{c}{Q}_{\text{total}}=448.29\text{ k´³}+237.95\text{ k´³}+55.59\text{ k´³}\\ =741.9\text{ k´³}\\ ≈742\text{ k´³}\end{array}$

Hence, the amount of heat to be removed is$742\text{ k´³}$