Question

At 1 atrn, how much energy is required to heat 770 g of H2O(s) at-120 ℃ to H2O(g) at 123.0 constants can be found here ? Helpful Number k.J

Answer 1

We require 2 type of heat, latent heat and sensible heat

Sensible heat (CP): heat change due to Temperature difference

Latent heat (LH): Heat involved in changing phases (no change of T)

Then

Q1 = m*Cp ice * (Tf – T1)

Q2 = m*LH ice

Q3 = m*Cp wáter * (Tb – Tf)

Q4 = m*LH vap

Q5 = m*Cp vap* (T2 – Tb)

Note that Tf = 0°C; Tb = 100°C, LH ice = 334 kJ/kg; LH water = 2264.76 kJ/kg.

Cp ice = 2.01 J/g°C ; Cp water = 4.184 J/g°C; Cp vapor = 2.030 kJ/kg°C

Then

Q1 = m*2.01 * (0 – T1)

Q2 = m*334

Q3 = m*4.184 * (100 – 0)

Q4 = m*2264.76

Q5 = m*2.03* (T2 – 100)

QT = m*(2.01 * (0 – T1)+334+4.184 *100 + 2264.76+2.03* (T2 – 100))

QT = 77*(2.01 * (0 – -12)+334+4.184 *100 + 2264.76+2.03* (123– 100))

QT = 237773.69 J

QT = 237.77 kJ