Question

Calculate the amount of heat (in kJ) that must be absorbed to convert 108 g of ice at `0^@"C"` to water at `70^@"C"`?

Answer 1

You'd need 67.6 kJ of heat to convert that much ice at `0^@"C"` to water at `70^@"C"`.

So, you need to go from ice at `0^@"C"`, which is still a solid, to water at `70^@C"`, which is of course a liquid. This implies that you will go through a phase change, i.e. from ice at `0^@"C"` to water at `0^@"C"`.

As a result, you're going to have to consider two heats, one needed for the phase change and the other needed to raise the temperature of water from `0` to `70^@"C"`.

The heat needed for the phase change will be

`q_1 = m * DeltaH_"fus"`, where

`m` - the mass of ice/water;
`DeltaH_"fus"` - the enthalpy of fusion.

Plug in your values to get

`q_1 = 108cancel("g") * 333"J"/cancel("g") = "35964 J"`

The heat needed to raise the temperature of the water will be

`q_2 = m * c_"water" * DeltaT`, where

`c_"water"` - the specific heat of water;
`DeltaT` - the change in temperature, defined as the difference between the initial and the final temperature of the water.

Plug in your values to get

`q_2 = 108cancel("g") * 4.18"J"/(cancel("g") ^@cancel("C")) * (70-0)^@cancel("C") = "31600.8 J"`

The total heat needed will be

`q_"total" = q_1 + q_2`

`q_"total" = 35964 + 31600.8 = "67654.8 J"`

I'll leave the answer rounded to three sig figs and expressed in kJ, so you'll get

`q_"total" = color(green)("67.6 kJ")`