I am doing a specific heat lab. I have 100 g of Al with the initial temperature

Question

I am doing a specific heat lab.

I have 100 g of Al with the initial temperature of 100 degree C.

I have 150 g of water at room temperature of 20 degree C.

When combining Al + water I get a final temperature of 34 degrees C.

I just need some help solving this equation:

(Mass metal)x(Specific heat of metal )x(Initial temp of metal - Final temp of metal) =
(Mass of water)x(Specific heat of water)x(Initial temp of water - Final temp of water)

Note: Final Temp of Water = Final Temp of Metal.

Note: (Specific heat of metal)(atomic weight) » 6 cal/mole, degree

I'm not quite sure how to solve this to find the specific heat and estimate of the atomic weight.

Explanation / Answer

heat lost by Al metal = heat gained by hot water

m*s*DT = m*s*DT

100*0.9*(100 - x) = 150*4.18*(x-20)

x = final temperature of mixer = 30.042 c

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