Question

How many grams of ice at -6.2 ∘C can be completely converted to liquid at 22.2 ∘C if the available heat for this process is 4.11×103 kJ ?

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

Answer 1

suppose mass = x

heat gained by ice to reach -6.2 C to 0 C

Q1 = mass * specific heat * change in temperature

= x* 2.01 * (0-(-6.2))

= 12.462 x J

moles of water = mass / molar mass

= x/18

heat gained to change state

Q2 = mole * fus

= x/18 * 6.01

= 0.334 x KJ or 334x

heat required to reach 0 to 22.2 C

Q3 = mass * specific heat * change in temperature

= x * 4.184 * (22.2 - 0)

= 92.885 x J

Q(total) = 4.11 * 10^3

12.462 x + 334 x + 92.885 x = 4.11 * 10^3

x = 9.355 grams

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