### Exercise 8.19

Posted:

**Sun Jan 21, 2018 7:27 pm**Does anyone know how to solve exercise 8.19 from the textbook?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=127&t=26024

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Posted: **Sun Jan 21, 2018 7:27 pm**

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

Posted: **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%

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%