Question

Question #16930

Answer 1

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))`