In: Chemistry
How much heat (in kJ) is required to warm 10.0 g of ice, initially at -10.0 ∘C, to steam at 114.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C, the heat of fusion for water is 6.02 kJ/mol, and the heat of vaporization for water is 40.7 kJ/mol.
Given data:
The specific heat capacity of ice (Cice) = 2.09 J/g⋅∘C
The heat of fusion for ice (Hf) = 6.02 kJ/mol
The specific heat capacity of water (Cwater) = 4.18 J/g⋅∘C
The heat of vaporization for water (Hv)= 40.7 kJ/mol
The specific heat capacity of steam (Csteam) = 2.01 J/g⋅∘C
Ice (s) Ice (s) water (l) steam (g) steam (g)
(-10 oC) (0 oC) (100 oC) (114 oC)
(T1) (T2) (T3) (T4)
We know that, the specific heat capacity of a substance,
C = Q/m(T2-T1), then Q = C*m*(T2-T1) (In general)
Here, Q = Cice*m*(T2-T1) + Hf*n + Cwater*m*(T3-T2) + Hv*n + Csteam*m*(T4-T3)
m = mass of ice, i.e. 10 g., n = no. of moles of ice, i.e. 10/18 = 0.556 mol, T2-T1 = 10, T3-T2 = 100 and T4-T3 = 14
Q = 2.09*10*10 + 6.02*1000*0.556 + 4.18*10*100 + 40.7*1000*0.556 + 2.01*10*14
= 209 + 3347.12 + 4180 + 22629.2 + 281.4
= 30646.72 J
= 30.65 KJ
Therefore, 30.65 KJ of heat is required to warm 10.0 g of ice, initially at -10.0 oC, to steam at 114.0 oC.