Question

Question #8a0c8

Answer 1

`"22.4 kJ"`

Explanation:

Your strategy here will be to

• determine how much heat is needed to heat the sample from ice at `-5.00^@"C"` to ice at `0^@"C"` by using the specific heat of ice

• determine how much heat is needed to heat the sample from ice at `0^@"C"` to liquid at `0^@"C"` by using the heat of fusion for water

• determine how much heat is needed to heat the sample from liquid at `0^@"C"` to liquid at `25^@"C"` by using the specific heat of water

You will need to know

• the specific heat of ice `-> 2.09"J"/("g" ""^@"C")`

• the specific heat of water `-> 4.18 "J"/("g" ""^@"C")`

• the heat of fusion for water `-> DeltaH_f = 334"J"/"g"`

So, you need to make sure that you include the phase change underwent by water when going from solid at its melting point to liquid at its melting point.

As you know, the specific heat of a substance tells you how much heat is needed to increase the temperature of `"1 g"` of that substance by `1^@"C"`.

The equation that relates heat absorbed / lost and change in temperature looks like this

`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

So, how much heat would be needed to convert `"50.0 g"` of solid ice at `-5.00^@"C"` to solid ice at `0^@"C"`?

`q_1 = 50.0 color(red)(cancel(color(black)("g"))) * 2.09 "J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * [0 - (-5.00)]color(red)(cancel(color(black)(""^@"C")))`

`q_1 = "522.5 J"`

Now, phase changes always take place at constant temperature. The transition from solid ice at `0^@"C"` to liquid water at `0^@"C"` can be described using the equation

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

`q` - heat absorbed
`m` - the mass of the sample
`DeltaH_f` - the heat of fusion for that substance

In your case, the heat needed for this phase change will bee equal to

`q_2 = 50.0 color(red)(cancel(color(black)("g"))) * 334"J"/color(red)(cancel(color(black)("g"))) = "16,700 J"`

Finally, the heat needed to heat the sample from liquid at `0^@"C"` to liquid at `25^@"C"` will be

`q_3 = 50.0 color(red)(cancel(color(black)("g"))) * 4.18 "J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (25 - 0)color(red)(cancel(color(black)(""^@"C")))`

`q_3 = "5,225 J"`

The total heat needed to go from ice at `-5.00^@"C"` to liquid at `25^@"C"` will thus be

`q_"total" = q_1 + q_2 + q_3`

`q_"total" = "522.5 J" + "16,700 J" + "5,225 J"`

`q_"total" = "22,447.5 J"`

I'll leave the answer expressed in kilojoules and rounded to three sig figs, despite the fact that you only have two sig figs for the final temperature of the sample

`q_"total" = color(green)("22.4 kJ")`