The volume of a ball was measured at 500.0 cm3, and then mass was measured on this triple beam balance. what would be the most a
2 answers:
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Explanation:
The given data is as follows.
mass, m = 75 g
Specific heat of water = 4.18
First, we will calculate the heat required for water is as follows.
q =
=
= 8464.5 J/mol
= 8.46 kJ ......... (1)
Also, it is given that = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.
Now, we will calculate the heat of fusion as follows.
q =
=
= -46.32 kJ ......... (2)
Now, adding both equations (1) and (2) as follows.
8.46 kJ - 46.32 kJ
= -37.86 kJ
Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at to ice at
is -37.86 kJ.