how do I do this question?

Question

How much energy is required to change a 42g ice cube from at -11C to steam at 111C?

Answer 1

First heat it up to 0 deg C
Q=mCp(ice)(T2-T1) [solid water]
Q= 42 x 2.114 (0-(-11))
Q=980 J

Second melt it
Q=mCf
Q=42 x 333.55=
Q=14000 J

Third heat to 100 deg C
Q=mCp(water)(T2-T1)
Q=42 x 4.1813x(100-0)
Q=17600 J

Forth evaporate it
Q=mCf(latent heat of evaporatiom)
Q=42 x 2,270 =95300 J

and
finally heat it up to 111 degC
Q=mCp(vapor)(T2-T1) [water in gas state]
Q=42x 2.080(111-100)=
Q=1290 J
Now just add them all up

Qt=Q1+Q2+Q3+Q4+Q5


Some constants
Cp(solid) =2.114 J /(g K)
Cp(water) =4.1813 J /(g K)
Cp(gas) = 2.080 J /(g K)
Cf (fusion) =333.55 J/g [latent heat of fusion]
Cf(evaporation) =2,270 J/g [latent heat of evaporation]

Answer 2

SH ice = 0.5cal/g/°C.
LH melting ice = 80cal/g.
SH water = 1cal/g/°C.
LH vaporising water = 540cal/g.
SH steam = 0.5cal/g/°C.

Calculation...
Heating the ice to 0°C..
= 42g x 0.5cal/g/°C x 11°C ΔT = 231cal.
Ice at 0°C to water at 0°C...
= 42g x 80cal/g = 3,360cal.
Water at 0°C to 100°C...
= 42g x 1cal/g/°C x 100°C ΔT = 4,200cal.
Water at 100°C to steam at 100°C...
= 42g x 540cal/g = 22,680cal.
Steam at 100°C to 111°C...
= 42g x 0.5cal/g x 11°C ΔT = 231cal.

Total heat required = 231 + 231 + 22,680 + 4,200 + 3,360
= 26,922 cal. = 22.922kcal.

(Joules = calories x 4.184J/cal = 26,922cal x 4.184J/cal
112,642 J = 112.642kJ)