Question

Calculate the heat absorbed when 578 g of ice at −12.7°C melts and then the water is converted to steam at 205.0°C. The specific heat of ice is 2.03 J/(g°C),the heat of fusion of ice is 6.01 ×103 J/mol, the specific heat of water is 4.18 J/(g°C),the heat of vaporization of water is 4.07 × 104J/mol, and the specific heat of steam is 2.02 J/(g°C).

Answer 1

Ti = -12.7 oC

Tf = 205.0 oC

here

Cs = 2.03 J/g.oC

Heat required to convert solid from -12.7 oC to 0.0 oC

Q1 = m*Cs*(Tf-Ti)

= 578 g * 2.03 J/g.oC *(0--12.7) oC

= 14901.418 J

Hfus = 6.01KJ/mol =

6010J/mol

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 578.0/18.016

= 32.0826 mol

Heat required to convert solid to liquid at 0.0 oC

Q2 = n*Hfus

= 32.0826 mol *6010 J/mol

= 192816.3854 J

Cl = 4.18 J/g.oC

Heat required to convert liquid from 0.0 oC to 100.0 oC

Q3 = m*Cl*(Tf-Ti)

= 578 g * 4.18 J/g.oC *(100-0) oC

= 241604 J

Hvap = 40.7KJ/mol =

40700J/mol

Heat required to convert liquid to gas at 100.0 oC

Q4 = n*Hvap

= 32.0826 mol *40700 J/mol

= 1305761.5453 J

Cg = 2.02 J/g.oC

Heat required to convert vapour from 100.0 oC to 205.0 oC

Q5 = m*Cg*(Tf-Ti)

= 578 g * 2.02 J/g.oC *(205-100) oC

= 122593.8 J

Total heat required = Q1 + Q2 + Q3 + Q4 + Q5

= 14901.418 J + 192816.3854 J + 241604 J + 1305761.5453 J + 122593.8 J

= 1877677 J

= 1877.7 KJ

Answer: 1.88*10^3 KJ