Exactly 16.8 mL of water at 30.0 degrees celsius are added to a hot iron skillet....

Exactly 16.8 mL of water at 30.0 degrees celsius are added to a hot iron skillet. All of the water is converted into steam at 100 degrees celsius. The mass of the pan is 1.40 kg and the molar heat capacity of iron is 25.19 J/(mol*degree celsius). What is the temperature change of the skillet in degrees celsius?

Expert Solution

density of water = 1 g/mL
specific heat capacity of water, C = 4.186 J/goC
latent heat of vaporization of water LV = 2264.76 J/g
mass of water= volume * density = 16.8 mL*1 g/mL = 16.8 g

Heat released by water = heat released when water is converted from 30 oC to 100 oC + heat released to convert water to steam
                  = m*C*delta T + m*Lv
                  = 16.8 * 4.186 * (100-30) + 16.8 * 2264.76
                  = 4922.74 + 38047.97
                  = 42970.71 J

Mass of iron = 1.4 Kg = 1400 g
Molar mass of iron = 55.85 g/mol
number of moles of iron,n = mass/ molar mass
=1400 / 55.85
= 25.07 mol

heat absorbed by iron = 42970.71 J
use:
Q = n*C'* delta T
42970.71 = 25.07 * 25.19 * delta T
delta T = 68 oC
Answer: 68 oC

queen_honey_blossom answered 2 years ago

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