Question #440b7

Apr 10, 2017

30.73 KJ.

Explanation:

use these equations:
$q = m c \Delta T$ - for temperature change
$q = \Delta {H}_{\text{fus/vap" * "moles}}$ - for phase changes

also, use these values:
$\Delta {H}_{\text{fusion}}$ = 6.01 KJ/mol
(KJ needed per mole of water to change solid$\leftrightarrow$liquid)
$\Delta {H}_{\text{vaporization}}$ = 40.7 KJ/mol
(KJ needed per mole of water to change liquid$\leftrightarrow$gas)
Specific heat of water = 4.18 J/g*C

First, you have to find the amount of heat energy required for each step.

Step 1: ice at -10 degrees to ice at 0 degrees
$q = m c \Delta T$
q = (10)(2.09)(10)
q = 209 J = .209 KJ

Step 2: ice at 0 degrees to water at 0 degrees
$q = \Delta {H}_{\text{fus" * "moles}}$
10 g water = .556 mol water (divide my molar mass which is 18g)
q = (6.01)(.556)
q = 3.34 KJ

Step 3: water at 0 degrees to water at 100 degrees
$q = m c \Delta T$
q = (10)(4.18)(100)
q = 4180 J = 4.180 KJ

Step 4: water at 100 degrees to steam at 100 degrees
$q = \Delta {H}_{\text{vap"* "moles}}$
q = (40.7)(.556)
q = 22.63 KJ

Step 5: steam at 100 degrees to steam at 120 degrees
$q = m c \Delta T$
q = (10)(1.84)(20)
q = 368 J = .368KJ