Question

Question #1112a

Answer 1

You'd need 97.9 kJ of heat to melt that much ice at `0^@"C"`.

Explanation:

The first thing you need to notice is that you need to melt the ice at its melting temperature. This means that you're dealing with a phase change and all the supplied heat will go into converting solid ice at `0^@"C"` to liquid water at `0^@"C"`.

The equation that you'll use looks like this

`q = n* DeltaH_"fus"`, where

`n` - the number of moles of ice;
`DeltaH_"fus"` - the molar enthalpy of fusion.

Since the molar heat of fusion expresses the amount of heat required to convert 1 mole of solid ice to liquid water at `0^@"C"`, you're going to have to determine how many moles of ice are present in your sample.

You can write the number of moles of a substance as the ratio between mass and molar mass

`n = m/M_M`

In order to get the mass of ice, you can use its density and volume

`rho = m/V => m = rho * V`

In your case, you have

`320cancel("cm"^3) * "0.917 g"/(1cancel("cm"^3)) = "293.44 g"`

Water's molar mass is equal to 18.02 g/mol, which means that you get

`q = (293.44cancel("g"))/(18.02cancel("g")/cancel("mol")) * 6.009"kJ"/cancel("mol") = "97.85 kJ"`

Rounded to three sig figs, the answer will be

`q = color(green)("97.9 kJ")`

You need to supply 97.9 kJ of heat in order to convert 293.44 g of ice at `0^@"C"` to liquid water at `0^@"C"`.