Achieve W3&4 #9

[Alexis Shen 2G]

Hi everyone! I'm having trouble with this problem:

I'm a bit confused overall on how to start this problem. Does anyone have any hints to get me started?

[Barbara Soliman 1G]

We need to set their q values equal to each other. Since cooler water is absorbing heat, that q value will be positive and the q value for the warmer water will be negative.

q(cooler water)= -q(warmer water)
mc(delta T)= -(mc(delta T)
(410g)(4.184J/g⋅°C)(Tfinal - 25°C) = -(100g)(4.184J/g⋅°C)(Tfinal - 95°C)

Solve algebraically for Tfinal and you should get your answer!

[Alexis Shen 2G]

We need to set their q values equal to each other. Since cooler water is absorbing heat, that q value will be positive and the q value for the warmer water will be negative.

q(cooler water)= -q(warmer water)
mc(delta T)= -(mc(delta T)
(410g)(4.184J/g⋅°C)(Tfinal - 25°C) = -(100g)(4.184J/g⋅°C)(Tfinal - 95°C)

Solve algebraically for Tfinal and you should get your answer!

Thank you! That makes a lot of sense. I have one quick question, why do we set the q of each water equal to each other?

[Jessica Cornelia Hongarta 1G]

The energy (heat) released by the warmer water, is absorbed by the cooler water. This is why q(cooler water) + q(warmer water) = 0 and we can say that q(cooler water) = -q(warmer water). Think of the 9-) sign as the warmer water losing energy (heat). Hope this helps!