Homework Question (6th Edition) 8.19

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Homework Question (6th Edition) 8.19

Postby 804980216 » Sat Feb 09, 2019 4:27 pm

Calculate the heat that must be supplied to a 500.0g copper kettle containing 400.0g of water to raise temperature from 22 degrees Celsius to the boiling point of water, 100.0 degrees Celsius.
I understand that to solve this problem I need to use the Calorimetry forumula q= MCAT. In this case, we need to calculate the MCAT of both the 500.0g copper kettle and the 400.0g water seperately and then add together. According the the solutions manual, heat for the 500g kettle is q = (500g)(0.38J/gK)(78)....why is C = 0.38 and not 4.18??


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Re: Homework Question (6th Edition) 8.19

Postby Stevin1H » Sat Feb 09, 2019 5:54 pm

The specific heat capacity for copper and water are two different values. According to the table in the textbook, the specific heat capacity of copper is 0.38 J/g.C and the specific heat capacity of water is 4.184 J/g.C

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