Question

512 g of an unknown metal at a temperature of 15°C is dropped into a 100 g aluminum container holding 325 g of water at 98°C. A short time later, the container of water and metal stabilizes at a new temperature of 78°C. Identify the metal.

Short answer

The result indicates the specific heat of aluminium

Step-by-step solution

Definition of specific heat 

The amount of heat needs to raise up 1-degree Celsius temperature is known as specific heat

Finding the specific heat 

The relation between specific heat with metal, aluminium and water is

$c_m,m_m\times ∆T=(c_w\times m_w+c_{Al}\times m_{Al})∆T$

specific heat of metal, water and aluminium and temperature difference defines

$$\begin{gathered} c_m=\frac{(c_w\times m_w+c_{Al}\times m_{Al})∆T}{m_m\times ∆T} \\ {c}_m=\frac{(4190\times 0.325+900\times 0.1\left(98-78\right)}{0.518\times (78-15)}=900\frac{J}{KgK} \end{gathered}$$

The metal is aluminium as its specified heat is same