Question

a) How much heat is required to change a 31.3 g ice cube from ice at -10.6°C to water at 47°C? (if necessary, use cice=2090 J/kg°C and csteam= 2010 J/kg°C) b)How much heat is required to change a 31.3 g ice cube from ice at -10.6°C to steam at 110°C?

Answer 1

A)

First you want to raise the temp of the ice to 0*C;
use the specific heat of ice times the mass and the change in temp
(2090 J/kgC) (.0313kg)(10.6C)=693.42 J

Then you want to change the ice to water;
heat of fusion times mass
(3.33*10^5 J/kg)(.0313kg)= 10422.9 J

Then increase the temp of the water to 100*C
specific heat of water times mass time change in temp
(4186 J/kgC)(.0313kg)(100C)=13102.18 J

Change water to steam;
heat of vaporization times mass
(2.26*10^6 J/kg)(.0313kg)= 70738 J

increase temp of steam to 47*C
specific heat of steam times mass times change in temp
(2010 J/kgC)(.0313kg)(47C)= 2956.911 J

Add up the energy for each step to get the total energy equalling to 97.913 kJ

PART-B

First you want to raise the temp of the ice to 0*C;
use the specific heat of ice times the mass and the change in temp
(2090 J/kgC) (.0313kg)(10.6C)=693.42 J

Then you want to change the ice to water;
heat of fusion times mass
(3.33*10^5 J/kg)(.0313kg)= 10422.9 J

Then increase the temp of the water to 100*C
specific heat of water times mass time change in temp
(4186 J/kgC)(.0313kg)(100C)=13102.18 J

Change water to steam;
heat of vaporization times mass
(2.26*10^6 J/kg)(.0313kg)= 70738 J

increase temp of steam to 110*C
specific heat of steam times mass times change in temp
(2010 J/kgC)(.0313kg)(110C)= 6920.43 J

Add up the energy for each step to get the total energy equalling to 101.876 kJ