How much heat Q is required to change 19.0 g of ice at −1.00C to 19.0 g of steam at 128C ?

 

The heat required is the sum of all heat energy required in each phase.

Calculate heat required to change temperature of ice from -1°c to 0°c.

Using Q1=MC( ice)?T= 19g×2.09J/g°c× 1°c=39.71J

The next heat required is to change all ice to water at 0°c

Q2= m?Hf, ?Hf = 334J/g(constant)

Q2=19g×334J/g=6346J

The next heat is heat for needed to raise temperature of water to 100°c from 0°c

Q3=MC(water)?T=19g× 4.18J/g°C×100°c=7942J

The next phase is vapourization at 100°c whose heat is given by

Q4=m?Hv, ?Hv=2257J/g

Q4= 19g×2257J/g=42883J

The last phase is heat required to change temperature of steam from 100°c to 128°c

Q5=mc(steam)?T= 19g×2.09J/g°C×28°C= 1111.885J

The heat required is the sum of heat needed in all 5 phases

Q=Q1+ Q2+Q3+Q4+Q5= 58322.59J

Q=58.32kJ

Please see the attached file for the complete solution