Question

Calculate the amount of energy it will take to change 20.0 Grams of Water from -3 C to changing it all to vapor at sea level? Please help?

Please calculate the amount of energy it will take to change 20.0 Grams of Water from -3 C to changing it all to vapor at sea level. Show all your work for credit.

heat of fusion of water = 334 J/g

heat of vaporization of water = 2257 J/g

specific heat of ice = 2.09 J/g·°C

specific heat of water = 4.18 J/g·°C

Answer 1

Energy is additive, so you can add up the heat required for the individual steps. ALWAYS outline this type of problem.

`overbrace("20.0 g ice")^(-3^@ "C") stackrel(Heat" ")(->) overbrace("20.0 g ice")^(0^@ "C") stackrel(Mel t" ")(->) overbrace("20.0 g water")^(0^@ "C")`

`stackrel(Heat" ")(->) overbrace("20.0 g water")^(100^@ "C") stackrel(Vapo rize" ")(->) overbrace("20.0 g steam")^(100^@ "C")`

We label these steps as `q_1`, `q_2`, `q_3`, and `q_4`, respectively, so that `q = q_1+q_2+q_3+q_4`.

The total heat flow involved is:

`q = "60.3 kJ"`

---

Anytime you heat or cool, `DeltaT ne 0`, so `q = mc_sDeltaT` will apply, where `m` is the `"g"` of substance and `c_s` is the specific heat capacity in `"J/g"^@ "C"`.

`q_1 = mc_(s,ice)DeltaT_(ice)`

`= "20.0 g" cdot "2.09 J/g"^@ "C" cdot (0^@ "C" - (-3^@ "C")) = ul"125 J"`

`q_3 = mc_(s,water)DeltaT_(water)`

`= "20.0 g" cdot "4.18 J/g"^@ "C" cdot (100^@ "C" - 0^@ "C") = ul"8360 J"`

Anytime you do a phase transition, you are at constant temperature and pressure.

Constant pressure means `q = nDeltaH`, and constant temperature occurs because all the heat is invested into changing the phase of the substance.

`n` is the mols of the substance, but since you gave values of `DeltaH` in `"J/g"`, we can use `m` instead for mass.

`q_2 = mDeltaH_(fus)`

`= 20.0 cancel"g ice" cdot "334 J"/cancel"g ice" = ul"6680 J"`

`q_4 = mDeltaH_(vap)`

`= 20.0 cancel"g water" cdot "2257 J"/cancel"g water" = ul"45140 J"`

As a result, the total heat involved is

`color(blue)(q) = q_1+q_2+q_3+q_4`

`= "125 J" + "6680 J" + "8360 J" + "45140 J"`

`=` `"60305 J"`

`=` `color(blue)("60.3 kJ")`