Question

Four ice cubes at exactly 0 ∘C with a total mass of 52.0 g are combined with 155 gof water at 90 ∘C in 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

Solution :-

Mass of ice = 52.0 g

Mass of water = 155 g

Temperature of the water = 90 C

When the ice is placed in water then at first the ice will start to melt and then its temperature will rise from 0 C to the final temperature

So to calculate the final temperature we can set up the equation as follows

Lets convert the heat of fusion from kJ/mol to J per g

(6.02 kJ*1000 J/ 1 kJ) *(1mol / 18.0156 g )= 334 J/g

-q hot water = q ice

-m*c*delta T = (m*delta Hfus) + (m*c*delta T)

-155 g * 4.18 J per g C *(Tf - 90 C) = (334 J/g*52.0g)+(52.0g*4.18 J per g C * (Tf-0 C))

-647.9 Tf +58311 = 17368 + 217.36 Tf

58311 – 17368 = 217.36 Tf + 647.9 Tf

40943 = 865.26 Tf

40943 / 865.26 = Tf

47.3 C= Tf

Therefore the final temperature is 47.3 C