How much energy (in kJ) is required to raise the temperature of 25.0 g of H2O from –25°C to 25°C ? (ΔH°fus of H2O = 6.02 kJ/mol ; molar heat capacity of solid H2O = 37.6 J/mol•°C ; molar heat capacity of liquid H2O = 75.4 J/mol•°C)
(a) 70.7 kJ
(b) 8.73 kJ
(c) 61.9 kJ
(d) 12.3 kJ
Ti = -25.0 oC
Tf = 25.0 oC
here
Cs = 37.6 J/mol.oC
Lets convert mass to mol
Molar mass of H2O = 18.016 g/mol
number of mol
n= mass/molar mass
= 25.0/18.016
= 1.3877 mol
Heat required to convert solid from -25.0 oC to 0.0 oC
Q1 = n*Cs*(Tf-Ti)
= 1.3877 mol * 37.6s J/mol.oC *(0--25) oC
= 1304.3961 J
Hvap = 6.02KJ/mol =
6020J/mol
Heat required to convert solid to liquid at 0.0 oC
Q2 = n*Hvap
= 1.3877 mol *6020 J/mol
= 8353.6856 J
Cl = 75.4 J/mol.oC
Heat required to convert liquid from 0.0 oC to 25.0 oC
Q3 = n*Cl*(Tf-Ti)
= 1.3877 mol * 75.4l J/mol.oC *(25-0) oC
= 2615.7305 J
Total heat required = Q1 + Q2 + Q3
= 1304.3961 J + 8353.6856 J + 2615.7305 J
= 12273.8122 J
= 12.3 KJ
Answer: d
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