Calorimetry: A 400-g piece of metal at 120.0°C is dropped into a cup containing 450 g of water at 15.0°C. The final temperature of the system is measured to be 40.0°C. What is the specific heat of the metal, assuming no heat is exchanged with the surroundings or the cup? The specific heat of water is 4190 J/(kg ∙ K).

Question

Quizplus · Accepted Answer

The specific heat of the metal can be calculated using the principle of conservation of energy, where the heat lost by the metal is equal to the heat gained by the water. The formula used is $q = mc\Delta T$, where $q$ is the heat transferred, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. Rearranging the formula to solve for the specific heat of the metal ($c_{metal}$) gives us $c_{metal} = \frac{m_{water}c_{water}\Delta T_{water}}{m_{metal}\Delta T_{metal}}$. Plugging in the given values, we find that the specific heat of the metal is approximately 1470 J/(kg ∙ K).

Calorimetry: a 400-G Piece of Metal at 120