Question

Part A: Find the amount of heat that must be extracted from 1.5 kg of steam at 130 ∘C to convert it to ice at 0.0 ∘C.

Part B: What speed would this 1.5-kg block of ice have if its translational kinetic energy were equal to the thermal energy calculated in part A?
a
TRUST THE PROCESS
a.) (this is the process you'll use to find the heat needed and the heat extracted)
1.ice at -20∘C → heated to ice at 0∘C
2.ice at 0∘C → heated to water at 0∘C
3.water at 0∘C → heated to water at 100∘C
4.water at 100∘C → heated to vapor (steam) at 100∘C
5.vapor (steam) at 100∘C → heated to vapor at 130∘C


1. m=1.5 Q=m(c)(delta T)
c(specific heat of ice)= 2090 =(1.5)(2090)(20∘C)
delta T = 0-(-20)= 20∘C =62700 J

2. m=1.5 Q=m(Lf)
Lf(fusion)= 33.5*10^4 =(1.5)(33.5*10^4)
=502500 J

3. m=1.5 Q=m(c)(delta T)
c(specific heat of water)= 4186 =(1.5)(4186)(100)
delta T= 100- 0= 100∘C =627900 J

4. m=1.5 Q=m(Lv)
Lv(vaporization)= 22.6*10^5 =(1.5)(22.6*10^5)
=3390000 J

5. m=1.5 Q=m(c)(delta T)
c(specific heat of vapor)=2010 =(1.5)(2010)(30)
delta T=130-100= 30∘C =90450 J

Now you are going to add all the J's up
62700 J + 502500 J + 627900 J + 3390000 J + 90450 J
Q =4673550 J


There ya go hope that helps and sorry not sure how to do part b...
a
kinda posted funky but hope that helped a lil