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A 1.50-kg iron horseshoe initially at 590

A 1.50-kg iron horseshoe initially at 590

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Answer #1

let the final temperature be T ,
here,
specific heat of iron = 0.444 J/g*C
specific heat of water = 4.186 J/g*c

here,
1500*0.444*(590-T) = 19000*4.186*(T - 26)
solving the equation ,
T = 30.68 degree C

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Answer #2

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Answer #3


FROM PRINCIPLE OF caloriemetry..
heat lost by iron = heat gained by water....

mi*Ci*(590-T) = mw*cw*(T-26)...

1.5*450 *(590-T) = 19*4179(T-26)...
2462676 = 80076 T...
final temparature T = 30.75 oC

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Answer #4

The heat energy out Q out = Q in

Q = m*c*?T

So 1.50kg*470J/kg-oC*(600-Tf) = 20.0kg*4186J/kg-oC*(Tf-25)

So (83720 + 705)*Tf = 423000 + 2093000 = 2516000

So Tf = 2516000/(84425) = 29.8oC

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Answer #5

Q out of the iron = Q into water

where Q = m*c*?T

so 1.50*470*(590 - Tf) = 19*4186*(Tf - 26)

so 705*(590 - Tf)=79534*(Tf - 26)

so 80239*Tf = 2067884
Therefore Tf = 2067884/80239* = 25.77 degree Celcius

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