Question

Calculate the amount of heat energy required to convert 15.0 g of water at 60.5^oC to steam at 124.0°C.

        (C_water = 4.18 J/g°C; C_steam = 2.02 J/g°C; molar heat of vaporization of liquid water = 4.07 ´ 10⁴ J/mol)

Answer 1

Heat required to change 15 grams of water from 60.5 °C to 100°C

Q1 = mass * specific heat of water *(∆T)

= 15*4.18*(100-60.5)

= 2476.65 joules

Now change 100°C water to 100°C steam

Moles of water = mass / molar mass

= 15/18

= 0.833

Q2 = ∆H(vap)*mole

Q2 = 4.07*10^4*0.833

= 33903.1 joules

Now change 100°C steam to 124°C steam

Q3 = mass * specific heat of steam * (∆T)

= 15*2.02*(124-100)

= 727.2 joules

Total heat = Q1 + Q2 + Q3

= 2476.65 + 33903.1 + 727.2

= 37106.95 joules

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