Exercise 8.19

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Rucha Kulkarni 2A
Posts: 43
Joined: Fri Sep 29, 2017 7:05 am

Exercise 8.19

Postby Rucha Kulkarni 2A » Sun Jan 21, 2018 7:27 pm

Does anyone know how to solve exercise 8.19 from the textbook?

Nora Sharp 1C
Posts: 53
Joined: Sat Jul 22, 2017 3:00 am

Re: Exercise 8.19

Postby Nora Sharp 1C » Sun Jan 21, 2018 7:38 pm

We need to calculate the temperature rise for copper and water separately because they have different specific heats. The specific heat gives us the amount of energy needed to raise a substance based on mass and temperature, so that's what we'll use:

Copper: 500g*(.38 J*C^-1*g^-1)*(100 degrees C - 22 degrees C) +
Water: 400g*(4.18 J*C^-1*g^-1)*(100 degrees C - 22 degrees C)

= 145236 J.

We can see this solution in the solutions manual as 1.45*10^5J, before significant figures come into play and it's changed to 1.4*10^2 kJ.

Find the percentage of heat that goes to water with this equation:

((heat transferred to water)/(total heat transferred to water and kettle))*100%


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