Question

Find the amount of heat supplied to decrease the volume of an ice water mixture by $1\;{\text{c}}{{\text{m}}^3}$ without any change in temperature. $\left( {{\rho _{ice}} = {\text{0}}{\text{.9, }}{{\text{L}}_{ice}} = 80\;{\text{cal/gm}}} \right)$
A) $360\;{\text{cal}}$
B) $500\;{\text{cal}}$
C) $720\;{\text{cal}}$
D) None of the above

Answer 1

In this question, the concept of the latent heat of fusion is used that is the amount of heat required to melt unit mass of ice without changing in the temperature. First, we discuss the significance of heat in a system. Discuss the effect of latent heat of fusion Calculate the volume decrease in ice with and from them find out with its mass. Finally calculate the heat supplied with the concept of latent heat.

Complete step by step solution:
In this question, the decrease in the volume of an ice mixture by $1\;{\text{c}}{{\text{m}}^3}$ and the other given data are ${\rho _{ice}} = {\text{0}}{\text{.9, }}{{\text{L}}_{ice}} = 80\;{\text{cal/gm}}$
As the relative density is given, from this we can calculate the density of ice as,
$\rho = {\rho _{ice}} \times {\rho _w}$.
We know that density of water is ${\rho _w} = 1000\;{\text{kg/}}{{\text{m}}^{\text{3}}}$ or $1\;{\text{g/c}}{{\text{m}}^3}$.
Now, we substitute the value of the density of the water in the expression to obtain the density of the ice as,

$$\Rightarrow \rho = \left( {0.9} \right)\left( 1 \right) \\\ \Rightarrow \rho = 0.9\;{\text{g/c}}{{\text{m}}^{\text{3}}} \\\$$

Now, we calculate the mass of the ice melt by using the formula,
$m = \rho V$
Here, the volume of the ice water decreases by $1\;{\text{c}}{{\text{m}}^3}$. The amount of ice melt should be $V = 10\;{\text{c}}{{\text{m}}^3}$.
Now, substitute the given values in the above equation as,
$m = \left( {0.9} \right)\left( {10} \right)$
$\Rightarrow m = 9\;{\text{g}}$
It is given that the latent heat of fusion of ice is $80$ calories per gram.
We know that, the latent heat of fusion is the amount of heat required to melt unit mass of ice without changing in the temperature,
So, we calculate the amount of heat transfer as,
$Q = mL$
Now, we substitute the values as,
$\Rightarrow Q = \left( 9 \right)\left( {80} \right)$
$\therefore Q = 720\;{\text{cal}}$
Thus, the amount of heat transfer is $720\;{\text{cal}}$.

Hence, the correct option is C.

Note: As we know that Latent heat of substance is associated with phase change of matter. The latent heat of vaporization is $540$ calories per gram. It is absorbed during vaporization and given up during condensation. For substances having phase transform from solid to gas and reverse it is called latent heat of sublimation.