The heat energy absorbed by 100.0 g of water is 4.18 kiloJoules.
Given:
The 100.0 g of water was heated from 20.0°C to 30.0°C.
To find:
The kilojoules of heat energy are absorbed when 100.0 g of water.
Solution:
Mass of water = m = 100.0 g
The specific heat capacity of the water = c = 4.184J/g°C
The initial temperature of the water = [tex]T_1=20.0^oC[/tex]
The final temperature of the water = [tex]T_2=30.0^oC[/tex]
The amount of heat absorbed by the water = Q
The heat required to change the temperature of 100.0 grams of water from 20.0°C to 30.0°C will be given by :
[tex]Q=m\times c\times (T_2-T_1)\\\\Q=100.0g\times 4.184J/g^oC\times (30.0^oC-20.0^oC)\\\\Q=4,184 J\\\\1 J = 0.001 kJ\\\\4,184 J=4,184\times 0.001 kJ\\\\=4.184 kJ\approx 4.18kJ[/tex]
The heat energy absorbed by 100.0 g of water is 4.18 kiloJoules.
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