Question

Question #34abf

Answer 1

`"7000 cal"`

Explanation:

The idea here is that you need to take into account four distinct heats

• the heat required to turn `"10 g"` of ice at its normal melting point of `0^@"C"` to liquid water at `0^@"C"`
• the heat required to heat `"10 g"` of liquid water from `0^@"C"`to its normal boiling point of `100^@"C"`
• the heat required to convert `"10 g"` of liquid water at `100^@"C"` to vapor at `100^@"C"`
• the heat required to heat `"10 g"` of steam from `100^@"C"` to `110^@"C"`

In your case, you will have

`10 color(red)(cancel(color(black)("g ice"))) * overbrace("80 cal"/(1color(red)(cancel(color(black)("g ice")))))^(color(blue)("the latent heat of fusion of water")) = "800 cal"`

This represents the heat needed to convert your sample from ice at `0^@"C"` to liquid water at `100^@"C"`, or `q_1`.

Now, in order to figure out how much heat is needed to heat the liquid water from `0^@"C"` to `100^@"C"`, you need to know the specific heat of water, which you can take as

`c_"water" = "1 cal g"^(-1)""^@"C"^(-1)`

In your case, you will need

`10 color(red)(cancel(color(black)("g"))) * overbrace("1 cal"/(1color(red)(cancel(color(black)("g"))) * 1^@"C"))^(color(blue)("the specific heat of water")) = "10 cal"""^@"C"^(-1)`

to increase the temperature of the sample by `1^@"C"`, which implies that in order to increase its temperature by `100^@"C"`, you will need

`100 color(red)(cancel(color(black)(""^@"C"))) * overbrace("10 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 10 g of water")) = "1000 cal"`

This represents the total heat need to increase the temperature of `"10 g"` of liquid water from `0^@"C"` to `100^@"C"`, or `q_2`.

Next, you will have

`10 color(red)(cancel(color(black)("g"))) * overbrace("540 cal"/(1 color(red)(cancel(color(black)("g")))))^(color(blue)("the latent heat of vaporization of water")) = "5400 cal"`

This represents the heat needed t convert `"10 g"` of liquid water from `100^@"C"` to vapor at `100^@"C"`, or `q_3`.

Finally, you can increase the temperature of `"10 g"` of steam by adding

`10 color(red)(cancel(color(black)("g steam"))) * overbrace("0.5 cal"/(1color(red)(cancel(color(black)("g steam"))) * 1^@"C"))^(color(blue)("the specific heat of steam")) = "5 cal"""^@"C"^(-1)`

which implies that in order to increase the temperature of `"10 g"` of steam by

`110^@"C" - 100^@"C" = 10^@"C"`

you will need

`10 color(red)(cancel(color(black)(""^@"C"))) * overbrace("5 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 10 g of steam")) = "50 cal"`

This represents the heat needed to increase the temperature of `"10 g"` of steam from `100^@"C"` to `110^@"C"`, or `q_4`.

You can thus say that you have

`q_"total" = q_1 + q_2 + q_3 + q_4`

which, in your case, is equal to

`q_"total" = "800 cal + 1000 cal + 5400 cal + 50 cal"`

`q_"total" = color(darkgreen)(ul(color(black)("7000 cal")))`

The answer must be rounded to one significant figure, the number of sig figs you have for the mass of ice.