4A.9 7th Edition
Posted: Sun Jan 27, 2019 6:59 pm
Any idea on how to calculate the final temperature of 50.7 g of water at 22 C when a block of Cu of 20g at 100C is placed in it?
Re: 4A.9 7th Edition
Posted: Sun Jan 27, 2019 7:56 pm
You have to plug given values into q = mcΔT and set heat(water) = -heat(copper).
For the first equation, heat(water) = 50.7g (4.18J/g°C)(Tf - 22°C)
For the second, heat(copper) = 20g (.38J/g°C)(Tf - 100°C)
We know that when the copper is placed in water, eventually the copper and water will reach the same temperature, so Tf in both these equations will be the same. Set heat(water) = -heat(copper) and solve for Tf. We set the heat(water) equal to the opposite of the heat(copper) because the heat lost by one will be gained by the other.