Question

Question #95b88

Answer 1

`"75,000 J"`

Explanation:

The trick here is to realize that you must keep track of two phase changes as you go from ice at `0^@"C"` to steam at `100^@"C"`.

More specifically, you have

• ice at `0^@"C"` to liquid water at `0^@"C"`
• liquid water at `0^@"C"` to liquid water at `100^@"C"`
• liquid water at `100^@"C"` to steam at `100^@"C"`

In order to be able to calculate the heat required to convert your sample, you need to know

• `DeltaH_"fus" = color(blue)("333.55 J g"^(-1)) ->` the enthalpy of fusion of water
• `c_"water" = color(purple)("4.18 J g"^(-1)""^@"C"^(-1)) ->` the specific heat of water
• `DeltaH_"vap" = color(darkorange)("2257 J g"^(-1)) ->` the enthalpy of vaporization of water

So, the first thing to calculate here is the heat required to go from ice at `0^@"C"` to liquid water at `0^@"C"`. To do that, use the enthalpy of fusion of water, which tells you the enthalpy change that accompanies the conversion of `"1 g"` of ice to liquid water at its normal melting point.

`25 color(red)(cancel(color(black)("g ice"))) * color(blue)("333.55 J"/(1color(red)(cancel(color(blue)("g ice"))))) = "8,338.75 J"`

Next, calculate the heat needed to heat your sample from liquid water at `0^@"C"` to liquid water at `100^@"C"`.

The speciifc heat of water tells you that you need `color(purple)("4.18 J")` of heat to increase the temperature of `color(purple)("1 g")` of liquid water by `color(purple)(1^@"C")`.

This means that in order to increase the temperature of `"25 g"` of water, you need

`25 color(red)(cancel(color(black)("g"))) * color(purple)("4.18 J")/(color(purple)(1)color(red)(cancel(color(purple)("g"))) * color(purple)(1^@"C")) = "104.5 J"^@"C"^(-1)`

So in order to increase the temperature of `"25 g"` of water by `1^@"C"`, you need to provide it with `"104.5 J"`. In your case, the temperature change is

`100^@"C" - 0^@"C" = 100^@"C"`

so you will need a total of

`100color(red)(cancel(color(black)(""^@"C"))) * "104.5 J"/(1color(red)(cancel(color(black)(""^@"C")))) = "10,450 J"`

Finally, to convert your sample from liquid water at `100^@"C"` to vapor at `100^@"C"`, use the enthalpy of vaporization of water.

In this case, you know that you need `color(darkorange)("2257 J")` of heat to convert `color(darkorange)("1 g")` of water from liquid to vapor at its normal boling point, which means that your sampel will need

`25 color(red)(cancel(color(black)("g"))) * color(darkorange)("2257 J"/(1color(red)(cancel(color(darkorange)("g"))))) = "56,425 J"`

Finally, add all the three heats to get the total heat needed

`"total heat" = "8,338.75 J" + "10,450 J" + "56,425 J"`

`color(darkgreen)(ul(color(black)("total heat = 75,000 J")))`

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the two temperatures.