Given 355 mLmL of water at 24 ∘C∘C, how much ice (in grams) would have to be added to this water so that after the ice has melted, the overall temperature of the system has been lowered to 5 ∘C∘C? (Assume the density of water is 1.0 g/mLg/mL.)

Mass of ice required = 79.4 g

Explanation:

Mass of water = density of water * volume of water

mass of water = 355 mL * 1.0 g/mL = 355 g

Quantity of heat emitted by water during the process of the melting of ice is given as, Q = mcΔT

where c is specific heat capacity of water = 4.18 J/g/°C

ΔT is temperature change = final temperature - initial temperature

m is mass of water

Q = 355 g * 4.18 J/g/°C * (5 - 24)°C

Q = -28194.1 J

Let the mass of ice that would absorb this heat be m

Total heat absorbed by the ice = heat  absorbed in fusion of ice at 0°C to water at 0°c + heat absorbed in changing from water at 0°C to water at 5°C

Heat  absorbed in fusion of ice at 0°C to water at 0°c, H₁ = mL

where m is mass of ice; L is latent heat of fusion of ice = 334 J/g

H₁ = m * 334 J/g

H₁ = (334 J/g)m

Heat absorbed in changing from water at 0°C to water at 5°C, H₂ = mcΔT

H₂ = m * 4.1 J/g/°C * (5 - 0)°C

H₂ = (20.9 J/g)m

Total heat absorbed = (334 J/g)m + (20.9 J/g)m = (354.9 J/g)m

Since heat evolved by water = heat absorbed by ice

28194.1 J = (354.9 J/g)m

m = 28194.1 J/354.9 J/g

m = 79.4 g

Therefore, mass of ice required = 79.4 g

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