Question

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.

Answer 1

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^o`C 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`