Question #16930

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Question #16930

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Answer 1 · 2016-03-09T03:03:54

I found #12g#

Explanation:

Here you could use the expression for the heat #Q# exchanged as:
#Q=mC_pDeltaT#
that with your data gives us:
#216=m*0.9*(35-15)#
so that:
#m=(216)/(0.9*20)=12g#

I think you have for aluminium:
#C_p=0.9J/(gcolor(red)(°C))#

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Question #16930

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