Calculate the amount of energy in kilojoules needed to change 207.0 g of water i

Question

Calculate the amount of energy in kilojoules needed to change 207.0 g of water ice at -10.0 degree C to steam at 125.0 degree C. The following constants for water may be helpful. C_p, ice = 36.39 J mol^-1 degree C^-1 C_p, liquid = 75.375 Jmol^-1 degree C^-1 C_p, steam = 37.11 J mol^-1 degree C^-1 delta H fus = 6.02 kJ mol^-1 delta H vap = 40.7 kJ mol^-1 Comment: notice the unit on the specific heat values. It uses 'per mole' rather than per gram.' This means the either (l) we have to change the specific heat values to the 'per gram' value (do this by dividing by the molar mass of water or (2) converting the grams of water to moles of water (by dividing by the molar mass of water).

Explanation / Answer

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)

QT = Q1+Q2+Q3+Q4+Q5

Note that all should be either converted to mass or moles... I prefer working with mass so:

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 = Q1+Q2+Q3+Q4+Q5

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

QT = m(2.01*(-T1) + 334 + 4184 + 2264.76 + 2.03*T2 - 2030 )

QT = 207*(2.01*(--10) + 334 + 4184 + 2264.76 + 2.03*125- 2030 )

Qt = 1040508.27 J

QT = 1040.508 kJ

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.