How much heat (energy) is released, when 2.00 moles (36.0 g) of water at 25.0°C is frozen, and then cooled to a temperature of -12.0°C?
Update:
Details:
Specific heat capacity of ice: 2.05 Jg^-1 K^-1
Specific heat capacity of water: 4.18 Jg^-1 K^-1
Specific heat capacity of steam: 2.08 Jg^-1 K^-1
Molar heat of fusion for H2O: 6.02 kJ mol^-1
Molar heat of vaporization for H2O: 40.7 kJ mol^-1
Melting point of ice: 0.00°C
Boiling point of water: 373 K
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Verified answer
Heat transfer.....
The water goes through three transitions.... You can easily get the answer with one chain calculation.
q = mcΔT(liquid water) + mHf + mcΔT(ice)
q = 36.0g(4.18J/gC x 25.0C + 334J/g + 2.05J/gC x 12.0C)
q = 16671.6 J ..... this is rounded to 3 significant digits .... 16,700J... or... 16.7 kJ... or... 8.33 kJ/mol
You will need to do 3 calculations and then add those answers together.
1. calculate heat released when 36.0 g water cools from 25C to 0 C:
q = m c (T2-T1)
q = 36.0 g (4.18 J/gC) (0 - 25 C) = -3762 J = -3.76 kJ
2. Calculate heat released when 2.00 moles of water freezes. (Note: The heat of fusion is written as a positive number, but when water freezes, heat is released, and so this will also have a negative value:
q = - Molar heat of fusion (moles)
q = -6.02 kJ/mol X 2.00 mol = -12.04 kJ
3. Calculate heat released when the ice formed cools from 0C to -12 C:
q = m c (T2-T1)
q = 36.0 g (2.05 J/gC) (-12-0 C) = -885.6 J = -0.886 kJ
Total heat lost = -16.6 kJ