Question

How many grams of ice at -23.1∘ C can be completely converted to liquid at 12.6∘C if the available heat for this process is 4.19×103 kJ ? For ice, use a specific heat of 2.01 J/(g⋅∘C) and ΔHfus =6.01kJ/mol.

Answer 1

Q = heat change for conversion of ice at -23.1 ^oC to ice at 0 ^oC + heat change for conversion of ice at 0^oC to water at 0^oC + heat change for conversion of water at 0^oC to water at 12.6 ^oC

Q = mcdt + mL + mc'dt'

Where

m = mass of ice = ?

c' = Specific heat of water = 4.186 J/g degree C

c = Specific heat of ice= 2.01 J/g degree C

L= Heat of fusion of ice = 6.01 kJ/mol = 6.01x103 J/mol = 6.01x103 J/molx(1 mol / 18 g )

= 333.9 J/g

dt' = 12.6 -0 =12.6 ^oC

dt = 0-(-23.1)=23.1 ^oC

Q = 4.19x10³ kJ = 4.19x10⁶ J

Plug the values we get

Q = m(cdt + L + c'dt' )

m = 9.675x10³ g

Therefore the required mass of ice is 9.675x10³ g