Answer
See answers.
Work Step by Step
The heat absorbed by all three substances is given by equation 14–2, $Q=mc\Delta T$. Solve for the mass, $m=\frac{Q}{c\Delta T}$.
$$m_{Cu}:m_{Al}:m_{water}=\frac{Q}{c_{Cu}\Delta T}:\frac{Q}{c_{Al}\Delta T}:\frac{Q}{c_{water}\Delta T}$$
In this problem, the heat and temperature rise are the same for all three materials.
$$ =\frac{1}{c_{Cu} }:\frac{1}{c_{Al} }:\frac{1}{c_{water} }$$
$$ =\frac{1}{390 }:\frac{1}{900 }:\frac{1}{4186 }$$
$$ =10.7:4.65:1$$
