Explanation:
The amount of heat [tex]Q[/tex] absorbed in the temperature variation of a material is:
[tex]Q=m. c. \Delta T[/tex] (1)
Where:
[tex]m=435g[/tex] is the mass of water
[tex]c[/tex] is the specific heat of the element. In the case of water [tex]c=1 cal/g\°C [/tex]
[tex]\Delta T[/tex] is the variation in temperature, which in this case is [tex]\Delta T=100\°C-25\°C=75\°C[/tex]
(The boiling point of water at the pressure of 1 atm is [tex]100\°C[/tex])
Rewriting equation (1) with the known values:
[tex]Q=(435g)(1 cal/g\°C)(75\°C)[/tex]
(2)
[tex]Q=32625 cal[/tex] (3)
Nevertheless, we are asked to find this value in Joules. So, we have to convert this 32625 calories to Joules, knowing the following:
[tex]1 cal=4.187 J[/tex] (4)
Hence:
[tex]Q=32625 cal=136600.875 J=136.6kJ[/tex]
