Question

How much ice (in grams) must melt to lower the temperature of 351 mL of water from 26 ∘C to 4 ∘C ? (Assume the density of water is 1.0 g/mL and that the ice is at 0 ∘C, the heat of fusion of ice is 6.02 kJ/mol.)

Answer 1

Volume of water = 351 mL

density of water = 1.0 g/mL

Hence, mass of water, m = 351 g

Given that temeperature of the water to be lowered from 26^∘C to 4^∘C.

So, temprature difference dT =26^∘C - 4^∘C = 22^∘C

Heat lost by water Q = mc dT

= (351 g) x (4.184 J/g/^oC) x (22^∘C)

= 32308.85 J

Heat lost by water Q = 32308.85 J

This heat is used to melt the ice.

Given that heat of fusion of ice =  6.02 kJ/mol = 6020 J/mol

This means, 6020 J heat is required to melt 1 mol of ice.

so, 32308.85 J heat is required to melt ? mol of ice.

? = (32308.85 J/ 6020 J) x 1mol ice

= 5.367 mol ice

Moles of ice = 5.367 mol

1 mol of ice = 18 g

Mass of 5.367 mol ice = 5.367 x 18 g = 96.6 g

Hence, mass of ice to be melt = 96.6 g

Therefore, 96.6 grams of ice must melt to lower the temperature of 351 mL of water from 26^∘C to 4^∘C.