Question

Question #e02ea

Answer 1

`"350 J"`

Explanation:

Before doing anything else, make sure that you understand what happens when a solid melts.

The first thing to take note of here is that since melting is a phase change, it must take place at constant temperature.

As you know, pure water melts and freezes at `0^@"C"`. This means that a solid `->` liquid phase change takes place when you're going from solid at `0^@"C"` to liquid at `0^@"C"`.

Notice that your sample of ice starts at a temperature of `-1.1^@"C"`. This means that before you can melt it, you must first heat it from solid at `-1.1^@"C"` to solid at `0^@"C"`.

This means that you will have to know two things

• the specific heat of ice, which tells you how much heat is needed to increase the temperature of `"1 g"` of ice by `1^@"C"`

• the enthalpy of fusion for water, which tells you how much heat is needed in order to convert `"1 g"` of ice from solid to liquid at its melting point

You can find these values here

`c_"ice" = 2.09"J"/("g" ""^@"C")" "`

`DeltaH_"fus" = 334"J"/"g"`

http://www.engineeringtoolbox.com/latent-heat-melting-solids-d_96.html

Two equations will come in handy here

`color(blue)(q = m * c * DeltaT)" "`, where

`q` - heat absorbed / lost
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as final temperature minus initial temperature

`color(blue)(q = m * DeltaH_"fus")" "`, where

`q` - heat absorbed / lost
`m` - the mass of the sample
`DeltaH_"fus"` - the enthalpy of fusion of the substance

So, to get `"3.5 g"` of ice to go from solid at `-1.1^@"C"` to solid at `0^@"C"`, you need to provide it with

`q_1 = 3.5 color(red)(cancel(color(black)("g"))) * 2.09"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * [(0 - (-1.1)]color(red)(cancel(color(black)(""^@"C"))) = "8.05 J"`

To get the solid ice from solid at `0^@"C"` to liquid at `0^@"C"`, you'll need

`q_2 = 3.5 color(red)(cancel(color(black)("g"))) * 334"J"/color(red)(cancel(color(black)("g"))) = "337.5 J"`

The total heat needed to go from ice at `-1.1^@"C"` to liquid water at `0^@"C"` will thus be

`q_"total" = q_1 + q_2`

`q_"total" = "8.05 J" + "337.5 J"`

`q_"total" = "345.55 J"`

Rounded to two sig figs, the answer will be

`q_"total" = color(green)("350 J")`

To get the answer in kilojoules, use the conversion factor

`"1 kJ" = 10^3"J"`

`q_"total" = 350 * 10^(-3)"kJ" = color(green)("0.35 kJ")`