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