Question

How much heat in kilojoules is required to warm 10.0 g of ice, initially at -10.0 ∘C, to steam at 125 ∘C. The heat capacity of ice is 2.09 J/g∘C and that of steam is 1.84 J/g∘C.​

Answer 1

mass of ice = 10.0 g

moles of ice = 10 / 18.02 = 0.555

-10.0 °C ----------> 0.0 °C ------------> 100.0 °C -------------> 125.0 °C

                 1        2       3             4              5

in this process 5 conversions are required.

Q1 = m Cp dT = 10 x 2.09 x (10) = 209 J

Q2 = n x delta H fusion = 0.555 x 6.01 x 10^3 = 3335.55 J

Q3 = m Cp dT = 10 x 4.18 x (100 ) = 4184 J

Q4 = n x delta H vap = 0.555 x 40.7 x 10^3 = 22588.5 J

Q5 = m Cp dT = 10 x 1.84 x (125 - 100) = 460 J

total heat = Q1 + Q2 + Q3 + Q4 + Q5

   = 30777 J

total heat = 30.8 kJ