Question

Assume 12,500 J of energy is added to 2.0 moles of `H_2O` as an ice sample at `0^@"C"`. The molar heat of fusion is 6.02 kJ/mol. The specific heat of liquid water is 4.184 J/g K. The remaining sample consists of?

Answer 1

The remaining sample consists of liquid water at `3.1^@"C"`.

So, you know that you're dealing 2.0 moles of water as ice at `0^@"C"`. Moreover, you know that you have a certain amount of heat, 12,500 J, to supply to the sample.

The first thing you need to determine is whether or not that much energy is enough to melt all the ice, i.e. to make the entire sample undergo a phase change.

Since water goes from solid to liquid at constant temperature, more specifically at `0^@"C"`, you know that

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

`n` - the number of moles of water you have;
`DeltaH_"fus"` - the molar enthalpy of fusion.

Plug in your values to get

`q_1 = 2cancel("moles") * 6.02"kJ"/cancel("mol") = "12.04 kJ"`

Expressed in Joules, you have

`q_1 = 12.04cancel("kJ") * "1000 J"/(1cancel("kJ")) = "12040 J"`

This means that all the ice has melted, and whatever energy you still have will now go into heating the liquid water.

Now, the specific heat of liquid water is actually equal to

`c_"water" = 4.18"J"/(g ^@"C")`

You need to convert this to the molar heat capacity of water, which represents the amount of heat required to raise the temperature of 1 mole of water by `1^@"C"`.

To do that, use water's molar mass

`4.18"J"/(cancel("g") ^@"C") * (18.02cancel("g"))/"1 mole" = 75.3"J"/("mol" ^@"C")`

Since you don't know how much the temperature of the water will increase by using your remaining energy, you can say that the final temperature is equal to `T_f`, so

`q_2 = n * C_"water" * DeltaT`, where

`n` - the number of moles of water you have;
`C_"water"` - the molar heat capacity of water;
`DeltaT` - the change in temperature.

After all the ice melted, you are left with

`q_2 = 12500 - 12040 = "460 J"`

This many Joules will get your water to a temperature of

`460cancel("J") = 2.0cancel("moles") * 75.3cancel("J")/(cancel("mol") ^@cancel("C")) * (T_f - 0)^@cancel("C")`

`460 = 150.6 * T_f => T_f = 460/150.6 = 3.054^@"C"`

Rounded to two sig figs, the answer will be

`T_f = color(green)(3.1^@"C")`

Therefore, the remaining sample consists of liquid water at `3.1^@"C"`.