Question

How much heat is required to heat 25 g of ice at -30°C to 70°C?

H2O(s) s = 2.0 J/(g°C)

H2O(l) s = 4.18 J/(g°C)

ΔH fus = 6.01 kJ/mol

Answer 1

given the ice is at -30 degree. we have to give heat to raise its temperature 70 degree and at 70 degree the ice will melt to water. so here the state of ice will change from solid to liquid.

now for calculating heat. there will be 3 step involved

1st is to convert ice at -30 to ice at 0 degree

for which we will calculate heat=mass* [H2O(s) s = 2.0 J/(g°C]*t, where t= 30 degree( -30 to 0 degree)

so heat1= 1500J

now for heat2, the ice will change to water, then heat is calculated as mass*[ΔH fus = 6.01 kJ/mol] = 8347.2J

make sure to convet kJ to J. BY multiplying 6.02 by 1000. and dividing it by mass of water 18gm as fusion H is given in kJ/mol. we have to convert it to J/gm.

now heat3 is given to convert water from 0 degee to 70 degree = mass*[H2O(l) s = 4.18 J/(g°C)]*t (t=70- 0=70) = 7315J

now for total heat add all the heats individually. answer is 17162.2J = 17.1622KJ.