Question

# 12.67 How much heat (in kJ) is required to warm 10.0 g of ice, initially at...

12.67

How much heat (in kJ) is required to warm 10.0 g of ice, initially at -13.0 ∘C, to steam at 112.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

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 = 10*2.01 * (0 – -13) = 261.3

Q2 = 10*334 = 3340

Q3 = 10*4.184 * (100 – 0) = 4184

Q4 = 10*2264.76 = 22647.6

Q5 = 10*2.03* (112– 100) = 243.6

QT = Q1+Q2+Q3+Q4+Q5 = 261.3+3340+4184+22647.6+243.6 = 30676.5 J

Q = 30676.5 J or = 30.67 kJ

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