During summers in India,one of the common practices to keep cool is to make ice balls of crushed ice,dip them in flavoured sugar syrup,and sip them. For this,a stick is inserted into crushed ice,and it is squeezed in the palm to make it into a ball. Equivalently in winter,in those areas where it snows,people make snowballs and throw them around. Explain the formation of a ball out of crushed ice or snow in the light of the $p-T$ diagram of water.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The $p-T$ phase diagram of water shows that the fusion curve (the boundary between solid and liquid phases) has a negative slope. This means that for water,an increase in pressure at a constant temperature near $0^{\circ}C$ lowers the melting point of ice.
When crushed ice or snow is squeezed in the palm,the pressure applied at the contact points between the ice particles increases. Due to the negative slope of the fusion curve,this increased pressure causes the ice to melt locally,even if the temperature is slightly below $0^{\circ}C$.
As the pressure is released (when the hand is opened or the pressure is redistributed),the local pressure drops back to atmospheric pressure. Consequently,the water that was formed by melting refreezes,acting as a binder that fuses the individual ice particles together into a solid,stable ball.

Explore More

Similar Questions

Heat is being supplied at a constant rate to a sphere of ice which is melting at the rate of $0.1 \, g/s$. It melts completely in $100 \, s$. The rate of rise of temperature thereafter will be ........ $^\circ C/s$ (Assume no loss of heat.)

Difficult
View Solution

Work done in converting one gram of ice at $-10^{\circ}C$ into steam at $100^{\circ}C$ is ....... $J$

Difficult
View Solution

$A$ hunter fired a metallic bullet of mass $m \ kg$ from a gun towards an obstacle and it just melts when it is stopped by the obstacle. The initial temperature of the bullet is $300 \ K$. If $\frac{1}{4}$ th of the heat is absorbed by the obstacle,then the minimum velocity of the bullet is (Melting point of bullet $= 600 \ K$,Specific heat of bullet $= 0.03 \ cal \ g^{-1} {}^{\circ} C^{-1}$,Latent heat of fusion of bullet $= 6 \ cal \ g^{-1}$) (in $ms^{-1}$)

$2 \ kg$ of ice at $-20^{\circ} C$ is mixed with $5 \ kg$ of water at $20^{\circ} C$. What is the final amount of water in the mixture (in $kg$)? (Given: specific heat of ice $= 0.5 \ cal/g^{\circ} C$,specific heat of water $= 1 \ cal/g^{\circ} C$,latent heat of fusion of ice $= 80 \ cal/g$)

Difficult
View Solution

The temperature of a body is given by $T = 6t^2 + 4$. The rate of change of temperature with respect to time at $t = 10 \ s$ is ........ $K/s$. ($T$ is in $Kelvin$ and $t$ is in $seconds$)

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo