Question

Consider a glass of 212 mL of water at 25°C. Calculate the mass of ice at -15°C that must be added to cool the water to 10°C after thermal equilibrium is achieved. To find the mass of water use the density of water = 1.0 g/mL. g

Answer 1

Volume of water =212ml density =1 g/ml Mass of water= Volume* density =100*1= 100 gm

Heat to be removed from water to bring down the temperature from 25 deg.c to 10 deg.c =

mass* specific heat of water* temperature differene= 100*4.184*(25-10)=6279 joules

This heat to be supplied by heating ice -15 deg.c to water at 10 deg,c

let mass of ice =m

1. Heat to be added to ice to change the temperature from -10 deg.c to 0 deg.c

=m* specific heat of ice* temperature diference = m*2.06*10= 20.6m

2. Latent heat of fusion at 0 deg.c=335 j/g

Heat that needs to be added from m gram= m*335 Joules

3. Heat to be added from 0 deg.c to 10 deg.c =mass* specific heat* temperature difference= m*4.184*10=41.84m

total heat to be added =sum of 1+2+3= 20.6m+335m+41.84m= 6279 joules ( heat given by water at 25 deg.c)

397.44m =6279 joules

m= 15.79 gms