Question

How much heat (in kJ) is needed to convert 866 g of ice at −10.0°C to steam at 126.0°C? (The specific heats of ice, water, and steam are 2.03 J/g · °C, 4.184 J/g · °C, and 1.99 J/g · °C, respectively. The heat of fusion of water is 6.01 kJ/mol, the heat of vaporization is 40.79 kJ/mol.)

Answer 1

We know, total amount of heat needed (Q) = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Here,

Q₁ = Quantity of heat needed to raise the temperature of ice from -10^oC to 0^oC

Q₂ = Quantity of heat needed to convert ice to water

Q₃ = Quantity of heat needed to raise the temperature of water from 0^oC to 100^oC

Q₄ = Quantity of heat needed to convert water to steam

Q₅ = Quantity of heat needed to raise the temperature of steam from 100^oC to 126^oC

Mass of ice = 866 g

Moles of H₂O = 866 g / 18.02 g/mol = 48.05 mol

Therefore,

Q₁ = Mass x Specific heat of ice x Change in temperature = (866 g) x (2.03 J/g.^oC) x (10 ^oC) = 17579.8 J = 17.57 kJ

Q₂ = Moles x Heat of fusion of water = (48.05 mol) x (6.01 kJ/mol) = 288.78 kJ

Q₃ = Mass x Specific heat of water x Change in temperature = (866 g) x ( 4.184 J/g. ^oC) x ( 100 ^oC) = 362.33 kJ

Q₄ = Moles x Heat of vaporization of water = (48.05 mol) x ( 40.79 kJ/mol) = 1959.95 kJ

Q₅ = Mass x Specific heat of steam x Change in temperature = (866g) x ( 1.99 J/g.^oC) x (26 ^oC) = 44.80 kJ

Therefore,

Total quantity of heat needed (Q)

= (17.57 kJ) + ( 288.78 kJ) + ( 362.33 kJ ) + (1959.95 kJ) + (44.80 kJ)

= 2673.43 kJ.