Question

How much energy is required to change a 50-g ice cube from ice at −25°C to steam at 105°C? ___MJ

Answer 1

Heat required is given by:

Q = Q1 + Q2 + Q3 + Q4 + Q5

Q1 = Heat required from -25 C ice to 0 C ice

Q2 = Heat required for phase change

Q3 = Heat required from 0 C water to 100 C water

Q4 = Heat required for phase change

Q5 = Heat required from 100 C steam to 105 C steam

So,

Q = Mi*Ci*dT1 + Mi*Lf + Mi*Cw*dT2 + Mi*Lv + Mi*Cs*dT3

dT1 = 0 - (-25) = 25

dT2 = 100 - 0 = 100 C

dT3 = 105 - 100 = 5 C

Ci = Specfic heat of ice = 2090 J-kg/C

Cw = Specfic heat of water = 4186 J-kg/C

Cs = Specfic heat of steam = 2010 J-kg-C

Mi = mass of ice = 50 gm = 0.05 kg

Lf = latent heat of fusion = 3.34*10^5 J/kg

Lv = latent heat of vaporization = 2.25*10^6 J/kg

Using above values:

Q = 0.05*2090*25 + 0.05*3.34*10^5 + 0.05*4186*100 + 0.05*2.25*10^6 + 0.05*2010*5

Q = 153245 J = 0.153 MJ

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