Calculate the amount of heat required to convert 120 g of ice at -20 degrees Celsius to steam at 120 degrees Celsius? (Sp. heat H2O (s)= 2.09 J/gC , Sp. heat H2O(l)= 4.18 J/gC , Sp. H2O(g)=2.03 J/gC ; heat of fusion H2O=333 J/g , heat of vap H2O= 2260 J/g)

The answer is:30.9 kJ

I did this:
1) melt ice: (10g) (333J/g)= 3330J
2) warm liquid from -20C to 120C
(10g)(120-20)(4.18 J/jC)= 4180J
3)vaporize liquid to steam
(10 g)(2260J/g)= 22,600J
4) add three values to get total
3330J + 4180J + 22600J = 30.1 kJ
I'm aware I made many errors. I need someone to show each step done right.

Question

Calculate the amount of heat required to convert 120 g of ice at -20 degrees Celsius to steam at 120 degrees Celsius? (Sp. heat H2O (s)= 2.09 J/gC , Sp. heat H2O(l)= 4.18 J/gC , Sp. H2O(g)=2.03 J/gC ; heat of fusion H2O=333 J/g , heat of vap H2O= 2260 J/g)

The answer is:30.9 kJ

I did this:
1) melt ice: (10g) (333J/g)= 3330J
2) warm liquid from -20C to 120C
(10g)(120-20)(4.18 J/jC)= 4180J
3)vaporize liquid to steam
(10 g)(2260J/g)= 22,600J
4) add three values to get total
3330J + 4180J + 22600J = 30.1 kJ
I'm aware I made many errors. I need someone to show each step done right.

aaronq · Answer

in step 2, you wrote: "(10g)(120-20)(4.18 J/jC)= 4180J"

T2-T1 should be 120-(-20)... 

q=(10g)(120+20)(4.18 J/jC)


everything else looks right

anonymous · Answer

ok! That's what I was not sure of!! Thank you!