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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?

User LITzman
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1 Answer

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Heat required to raise the temperature of water is given as


Q = ms\Delta T

here we know that


m = mass = 200 mL = 0.200 kg

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


Q = 0.200 (4186) (40 - 15) = 20930 J


User Ekhanna
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