Question

A 53-g piece of an unknown metal is heated to 119°c and then placed into a calorimeter containing 165g of water at a temperature of 19.0°c.The final temperature of both is 32.0°c. What is the specific heat of the unknown metal?

Answer 1

These problems require considering the heat source and the water as its own equilibrium. Recall,

`q = mC_sDeltaT`

The heated metal will lose heat, and the water will gain heat.

So,

`-q_"M" = q_(H_2O)`

and by extension,

`-m_"M"C_"M"DeltaT_"M" = m_(H_2O)C_(H_2O)DeltaT_(H_2O)`

The problem states `T_"f"` for both is `32.0°"C"` once the system has equilibrated.

Hence,

`=> C_"M" = (-m_(H_2O)C_(H_2O)DeltaT_(H_2O))/(m_"M"DeltaT_"M") approx (1.94"J")/(g*°"C")`

is the specific heat of the metal given your data.