Question

How many grams of ice at -5.7 ∘C can be completely converted to liquid at 20.0 ∘C if the available heat for this process is 5.20×10³ kJ ?

For ice, use a specific heat of 2.01 J/(g⋅∘C)J and ΔHfus=6.01kJ/mol

Answer 1

specific heat of solid water(ice)= 2.01 J/g °C

specific heat of liquid water= 4.18 J/g°C

heat release when water chsnges from -5.7 °c to 0 °C

Q1 = m x Cp(ice) x Delta T

= m x 2.01 x (0-(-5.7))

= 11.457 m J

heat release due to fusion. ∆H= 334 J/g

Q2 = m x∆H

= 334 m J

heat release while move to 20°C from 0°C

Q3 = m x Cp(l) × ∆T

= m × 4.18 × 20

= 83.6 m J

total heat

Q= Q1 + Q2 +Q3

=429.057 m J

mass of water = (5.2 × 10^3)/429.057

= 12.12 g