#8.) If 500 g of brass at 200 degrees C and 300 g of steel at 150 degrees C are added to 900 g of water in an aluminum pan of mass 150 g both at 20 degrees C, find the final temp. (assuming no loss of heat to the surrounding) The book answer is Tf = 32.5949 degrees C So far I am to solve the equation this way: Qlost = Qgained I have my equation set up as: 9220 cal -46 cal/oC Tf + 34.5 cal/oC Tf = 18000 cal -900 cal/oC Tf + 660 cal -33 cal/oC Tf I can see that adding 9200 cakk + 5175 cal + 18000 cal + 660 cal = 33035 cal/oC and adding up 900+33+46+34.5 = 1013.5 , therefore dividing 33035 by 1013.5 = 32.5949 degrees C. However, I see that is proper way to solve Tf algebraically ... How do I get the answer showing my work the correct way ... Thanks :)
#8.) If 500 g of brass at 200 degrees C and 300 g of steel at 150 degrees C are added to 900 g of water in an aluminum pan of mass 150 g both at 20 degrees C, find the final temp. (assuming no loss of heat to the surrounding)
The book answer is Tf = 32.5949 degrees C
So far I am to solve the equation this way: Qlost = Qgained
I have my equation set up as:
9220 cal -46 cal/oC Tf + 34.5 cal/oC Tf = 18000 cal -900 cal/oC Tf + 660 cal -33 cal/oC Tf
I can see that adding 9200 cakk + 5175 cal + 18000 cal + 660 cal = 33035 cal/oC
and adding up 900+33+46+34.5 = 1013.5 , therefore dividing 33035 by 1013.5 = 32.5949 degrees C. However, I see that is proper way to solve Tf algebraically ... How do I get the answer showing my work the correct way ... Thanks :)
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
