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0.5000 kg of water at 35.00 degrees celsius is cooled, with the removal of 6.300 e4 j of heat. what is the final temperature of the water? specific heat capacity of water is 4186 j/(kg c°).

remember to identity all of your data, write the equation, and show your work.

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J  = - 63000 J

Cp = 4184 J / kg°C

Formula

                        Q = mCp(T2 - T1)

                         T2 = T1 + Q/mCp    

Substitution

                       T2 = 35 - 63000/(0.5 x 4184)

                        T2 = 35 - 63000/2092

                        T2 = 35 - 30.1

                         T2 = 4.9 °C

explanation:

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