Question

How much heat (in kJ) is required to convert 423 g of liquid H2O at 24.0°C into steam at 152°C? (specific heat of liquid water is 4.184 J/g·°C, the specific heat of steam is 2.078 J/g·°C, and that both values are constant over the given temp)?

How much heat (in kJ) is required to convert 423 g of liquid H2O at 24.0°C into steam at 152°C? (Assume that the specific heat of liquid water is 4.184 J/g·°C, the specific heat of steam is 2.078 J/g·°C, and that both values are constant over the given temperature ranges. The normal boiling point of H2O is 100.0°C. The heat of vaporization (ΔHvap) is 40.65 kJ/mol.) `Delta` `"H"_v` `=` `"2257 J/g"`.

Answer 1

It requires `"1040.000 kJ"`.

Explanation:

We will be using the equation `Q=mcDeltT`,

where:

`Q` is the energy lost or gained, `m` is the mass of the water, `c` is the specific heat, of which there will be two; one for water and one for steam, and `DeltaT` which is the final temperature minus the initial temperature.

There are four steps required to answer this question. First, we will determine `Q` of liquid water. Second we will determine `Q` required to convert the given mass of liquid water to steam. Then we will determine `Q` for the steam. Then we will add all values for `Q` to get the total amount of energy needed to convert the given mass of liquid water to steam. The value of `Q` will be in Joules, so we will have to convert `"J"` to `"kJ"` by dividing `"J"` by `"1000"`.

Known

`m="423 g H"_2"O"`

`c_(H_2O(l)` `=("4.185 J")/("g"*""^@"C")`

`c_(H_2O(g)` `=("2.078 J")/("g"*""^@"C")`

`Delta` `"H"_v` `="2257 J/g"`

`DeltaT_(H_2O(l)` `=(100.0^@"C"-"24.0"^@"C")="76.0"^@"C"`

`DeltaT_(H_2O(g)` `=(152^@"C"-"100.0"^@"C")="52.0"^@"C"`

Unknown

`Q_(H_2O(l))`

`Q_("phase change")`

`Q_(H_2O(l))`

STEP 1: Determining `Q` for liquid water.

`Q` `=` `(423color(red)cancel(color(black)("g")))(("4.185 J")/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C"))))(76.0^@color(red)cancel(color(black)("C")))="24166.784 J"`

STEP 2: Determining `Q` for phase change from liquid to steam.

`Q=mDeltaH_v"`

Plug in the known values and solve.

`Q=(423color(red)cancel(color(black)("g")))(2257"J"/color(red)cancel(color(black)("g")))="974711 J"`

STEP 3: Determining `Q` for temperature change from `"100.0"^@"C"` to `"152"^@"C"`

`Q=(423color(red)cancel(color(black)("g")))(("2.078 J")/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C"))))(52^@color(red)cancel(color(black)("C")))="45707.668 J"`

STEP 4: Add all 3 values for `Q`.

`Q_"total"="24166.784 J"` + `"974711 J"` + `"45707.668 J"` `=` `"1044585 J"` `=` `"1040.000 J"` (rounded to three significant figures)

Convert Joules to kilojoules.

`"1 kJ"` `=` `"1000 J"`

`1040000color(red)cancel(color(black)("J"))xx("1 kJ")/(1000color(red)cancel(color(black)("J")))="1040.000 kJ"`

Note: Since we are working with grams and not moles, I added `DeltaH_v="2257 J/g"` to the end of your question, and used that value in STEP 2 of the answer.