# Question 16930

Mar 9, 2016

I found $12 g$

#### Explanation:

Here you could use the expression for the heat $Q$ exchanged as:
$Q = m {C}_{p} \Delta T$
that with your data gives us:
$216 = m \cdot 0.9 \cdot \left(35 - 15\right)$
so that:
$m = \frac{216}{0.9 \cdot 20} = 12 g$

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