Asked by martha
A 12-kg iron ball is dropped onto a pavement from a height of 90m. Suppose half of the heat generated goes into warming the ball. What is the temperature increase of the ball in degree C
Answers
Answered by
Elena
mgh/2=mcΔT
For iron c=444 J/kg•K
ΔT=gh/2c = 9.8•90/444•2=0.99℃
For iron c=444 J/kg•K
ΔT=gh/2c = 9.8•90/444•2=0.99℃
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