Question

How many grams of ice at -22.6 ∘C can be completely converted to liquid at 29.9 ∘C if the available heat for this process is 5.47×103 kJ ? For ice, use a specific heat of 2.01 J g−1 ∘C−1 and ΔfusH=6.01kJ mol−1.

Answer 1

Suppose mass = x grams

Energy required to reach -22.6 °C to 0°C

Q1 = mass * specific heat of ice * ∆T

= x * 2.01 * ( 0-(-22.6)) = 45.426 x J

Moles of water = mass / molar mass

= x / 18

Energy required to change the phase

Q2 = moles * ∆H(fus)

= x/18 * 6.01 = 0.334 x KJ or 334 x J

Energy required to change 29.9°C

Q3 = x * 4.184 * ( 29.9 - 0 ) = 125.1 x J

Q (total) = Q1 + Q2 + Q3

5.47 * 10^3 = 45.426 x + 334 x + 125.1 x

x = 10.842 grams

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