Question

How much energy (in kJ) is required to raise the temperature of 25.0 g of H2O...

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

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Homework Answers

Answer #1

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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