Question

The amount of heat supplied to decrease the volume of an ice water mixture by $1{\text{cm}}^3$ without any change in temperature is-
$(\text{Take}{\rho }_{\text{ice}}=0.9{\rho }_{\text{water}},L_f=80{\text{cal g}}^{-1})$

• A. $360\text{cal}$
• B. $500\text{cal}$
• C. $720\text{cal}$
• D. None

Answer 1

The correct option is C $720\text{cal}$

Given, ${\rho }_{\text{ice}}=0.9{\rho }_{\text{water}};{\rho }_{\text{water}}=1{\text{g cm}}^{-3};L_f=80{\text{cal g}}^{-1}$

$\because {\rho }_{ice}<{\rho }_{water}$

Hence, the volume of system can be decreased by $1{\text{cm}}^3$ keeping the temperature constant, only if some ice melts to form water.

Let $x\text{g}$ ice converts to form $x\text{g}$ water,

$\Rightarrow \Delta V=1{\text{cm}}^3$

Initial volume of $x\text{g}$ ice,

$V_i=\frac{m}{\rho }=\frac{x}{0.9}$

Initial volume of $x\text{g}$ water,

$V_w=\frac{m}{\rho }=\frac{x}{1}$

$\Rightarrow \Delta V=V_i-V_w$

$1=\frac{10x}{9}-x$

$\frac{x}{9}=1$

$\therefore x=9\text{g}$

Hence, the amount of heat required is,

$Q=m{L}_f$

$Q=9\times 80=720\text{cal}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.

Why this question? Tip: In this problem it is given that temperature should not change, thus it gives hint for only phase change involved. Key concept: It can be logically thought that "since density of water is greater than ice ", hence conversion of ice to water will lead to reduction in volume, keeping the mass constant.