Question

Find the amount of heat supplied to decrease the volume of an ice water mixture by $1{\text{cm}}^3$ without any change in temperature.
(${\rho }_{\text{ice}}=0.9{\rho }_{\text{water}},L_{\text{ice}}=80\text{cal/gm})$

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

Answer 1

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

Ice changes to water, so volume decreases but mass remains same.
Hence
$V_w{\rho }_w=V_{\text{ice}}{\rho }_{\text{ice}}$
$V_w=\frac{V_{\text{ice}}{\rho }_{\text{ice}}}{{\rho }_w}$
Change in Volume
$\Delta V=V_{\text{ice}}-V_w=(V_{\text{ice}}-\frac{V_{\text{ice}}{\rho }_{\text{ice}}}{{\rho }_w})$
$V_{\text{ice}}(1-0.9)=0.1V_{\text{ice}}=1{\text{cm}}^3$
$V_{\text{ice}}=10{\text{cm}}^3$
Heat required to melt volume $\left(V_{\text{ice}}\right)$ to water is
$H=\left(0.9{\rho }_w{V}_{\text{ice}}\right)L....\left(1\right)$
So from eq. (1)
$H=[0.9\times 1\times 10]\times 80$
$H=720\text{cal}$