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Would anyone be able to help me with this? I add a 50. g piece of...

Would anyone be able to help me with this?

I add a 50. g piece of Al (c = 0.88 J/g-deg) that is at 225°C to 100. mL of water at 20°C. What is the final temperature of the water? The density of water is approximately 1g/mL.

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Answer #1

volume of water is 100 mL and its density is 1 g/mL
So,
m(water) = 100 g
T(water) = 20.0 oC
C(water) = 4.184 J/goC
m(Al) = 50.0 g
T(Al) = 225.0 oC
C(Al) = 0.88 J/goC
T = 25.6 oC
We will be using heat conservation equation

Let the final temperature be T oC
use:
heat lost by Al = heat gained by water
m(Al)*C(Al)*(T(Al)-T) = m(water)*C(water)*(T-T(water))
50.0*0.88*(225.0-T) = 100.0*4.184*(T-20.0)
44*(225.0-T) = 418.4*(T-20.0)
9900 - 44*T = 418.4*T - 8368
T= 39.5 oC
Answer: 39.5 oC

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