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Exercise 8.19

Posted: Sun Jan 21, 2018 7:27 pm
by Rucha Kulkarni 2A
Does anyone know how to solve exercise 8.19 from the textbook?

Re: Exercise 8.19

Posted: Sun Jan 21, 2018 7:38 pm
by Nora Sharp 1C
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%