When 200.0mL of water is heated from 15.0 deg Celsius to 40.0 deg Celsius, how much thermal energy is absorbed by the water?

Respuesta :

Heat required to raise the temperature of water is given as

[tex]Q = ms\Delta T[/tex]

here we know that

[tex]m = mass = 200 mL = 0.200 kg[/tex]

initial temperature = 15 degree C

final temperature = 40 degree C

specific heat capacity = 4186 J/kg C

now by the above formula we will have

[tex]Q = 0.200 (4186) (40 - 15) = 20930 J[/tex]


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