Question

Four ice cubes at exactly 0 ∘C with a total mass of 51.5 g are combined with 130 g of water at 85 ∘Cin an insulated container. (ΔH∘fus=6.02 kJ/mol, cwater=4.18J/g⋅∘C)

If no heat is lost to the surroundings, what is the final temperature of the mixture?

Answer 1

Ice cubes at 0 deg.c first gain latent heat of fusionto become liquid at 0 deg.c

Total mass of 4 ice cubes =51.5gm, since latent heat is given in the units of Kj/mol, the mass given will be converted to mol

Moles of ice= 51.5/18=2.86 moles

Total heat that needs to be removed due to latent heat= moles of ice* latent heat of fusion =2.86*6.02 Kj/mol=17.2Kj   (1)

Now let the temperature to which this liquid water is raised be T.

so heat gained by 51.5gm of liquid water in the form of sensible heat= mass*Specific heat*Temperature difference

= 51.5*4.18*(T-0)= 215.27 JoulesT= 0.22T Kj   (2)

Total heat taken by ice = 17.2+ 0.22T (3)

Since there is no heat loss or gain to the surroundings, This gain in heat has to come from loss of heat from 130 gms of water from 130 deg.c to T.

Heat lost by 130 gms of water ( Sensible heat)= 130*4.18*(85-T) =543.4*(85-T ) Joules= 0.54*(85-T) (4)

Eq.3 =Eq.4

17.2+0.22T= 0.54*(85-T)=0.54*85-0.54T

T= 37.67 deg.c