How many grams of ice at -13.1 ∘C can be completely converted to liquid at 26.5 ∘C if the available heat for this process is 5.52×103 kJ ?

For ice, use a specific heat of 2.01 J/(g⋅∘C) and ΔHfus=6.01kJ/mol.

1 Answer

The available heat can convert 11.7 kg of ice.

Explanation:

First, determine the heat needed to warm the ice to 0^oC, then how much heat you need to melt the ice, and then how much heat you need to warm the water.

Solve each in terms of m. Add them together and set the sum equal to the total energy, then solve for m.

The total energy is the energy needed to
warm the ice to 0^oC + melt ice + warm the water.

q_1 +q_2+q_3= 5.52xx10^3kJ or 5.52xx10^6J

For ice, ΔT= 13.1^oC; C = 2.01 J/g^oC
q_1 = mC ΔT
q_1= m(2.01 J/g^oC)(13.1^oC)
q_1= m(26.33 J/g)

q_2=mΔH_f

Convert ΔH_f to J/g
((6.01(kJ)/(mol))(1000J/(kJ)))/((18.01g)/(mol))
ΔH_f= 334J/g

q_2=m(334 J/g)

Since you start at zero, ΔT is 26.5^oC for heating water

q_3= mCΔT
q_3= m(4.18 J/g^oC)(26.5^oC)
q_3=m(110.77J/g)

q_1 + q_2 + q_3= 5.52xx10^3 kJ or 5.52xx10^6J

m(26.33 J/g) + m(334 J/g) + m(110.77J/g) = 5.52 xx 10^6 J

m(471 J/g)= 5.52 xx 10^6 J
m= (5.52 xx 10^6 J)/(471 J/g)
m= 11 700 g
m=11.7 kg