Question

A spoon (13.4g) at 25C is used to stir the hot tea (8.0oz and 95C). The temperature of spoon and the tea ended up at 89C. What is the specific heat capacity of the spoon? Use specific heat capacity of water for tea.

Answer 1

The 2nd Law of Thermodynamics is your best friend. You know that the tea "water" must cool down, and the spoon must heat up. You also know that they must transfer the same amount of energy\[Q_{spoon} = Q_{tea}\] You also know that the heat transferred can be found by using\[Q = m*C*\Delta T\] you need to use this equation not once, but twice.

Find the heat released by the tea as it cools (be sure to convert the 8oz. into grams).

That same amount of heat must go into the spoon. Solve the equation again, but use the mass and DT of the spoon to find the C of the spoon.\[Q_{spoon} = Q_{tea}\]so \[m_{spoon}*C_{spoon}*\Delta T_{spoon} = m_{tea}*C_{tea}* \Delta T_{tea}\]and \[C_{spoon} = \frac{m_{tea}*C_{tea}* \Delta T_{tea}}{m_{spoon}* \Delta T_{spoon}}\]