A 400-g piece of metal at 100°C is dropped into a cup containing 450 g of water at 15.0°C. The final temperature of the system is 40.0°C. What is the specific heat of the metal, assuming no heat is exchanged with the surroundings or the cup?

Question

Quizplus · Accepted Answer

The heat lost by the metal equals the heat gained by the water. Using the formula $m \cdot c \cdot \Delta T$, solve for the specific heat of the metal: $c = \frac{(m_{water} \cdot c_{water} \cdot \Delta T_{water})}{(m_{metal} \cdot \Delta T_{metal})}$. Plugging in the values gives the specific heat of the metal as 1960 J/(kg∙K).

A 400-G Piece of Metal at 100°C Is Dropped into a Cup