How much heat (in kJ) is required to warm 13.0 g of ice, initially at -14.0 ∘C, to steam at 110.0 ∘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 = 13*2.09 * (0 – -14) = 380.38
Q2 = 13*334 = 4342
Q3 = 13*4.184 * (100 – 0) = 5439.2
Q4 = 13*2264.76 = 29441.88
Q5 = 13*2.01* (110– 100) = 261.3
QT = Q1+Q2+Q3+Q4+Q5 = 380.38 +4342+5439.2 + 29441.88+261.3 = 39864.76 J = 39.86 kJ
How much heat (in kJ) is required to warm 13.0 g of ice, initially at -14.0...
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