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 .
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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