Question

You initially have 2.0kg of ice in 3.0kg of water at 0°C. How much total heat must be added in order to convert 2.0kg of the entire sample to steam at 100°C? Separately determine the amount of heat for each stage of this process.   

Specific heat capacities (J/kg∙K) Latent heats (J/kg)

c ice = 2090 Lf = 33.5∙10^4

c water = 4186   Lv = 22.6∙10^5

c steam = 2010

Answer 1

Heat required is given by:

Q = Q1 + Q2 + Q3 + Q4 + Q5

Q1 = Heat required for phase change from ice to water

Q2 = Heat required from 0 C water to 100 C water

Q3 = Heat required for phase change

So,

Q = Mi*Lf + (Mi + Mw)*Cw*dT1 + (Mi + Mw)*Lv

given, Mi = mass of ice = 2.0 kg

Mw = mass of water = 3.0 kg

dT1 = 100 - 0 = 100 degC

Lv = 22.6*10^5 J/kg

Lf = 33.5*10^4 J/kg

Cw = 4186

Using above given values:

Q1 = Mi*Lf = 2.0*33.5*10^4 = 67.0*10^4 J = 6.7*10^5 J

Q2 = (Mi + Mw)*Cw*dT1 = (2+3)*4186*100 = 209.3*10^4 J = 2.09*10^6 J

Q3 = (Mi + Mw)*Lv = (2 + 3)*22.6*10^5 = 1130*10^4 J = 1.13*10^7 J

So, Total heat required = Q = (67.0 + 209.3 + 1130)*10^4 = 1406.3*10^4 J

Q = 1.406*10^7 J

"Let me know if you have any query."