Calculate the final temperature of a 63.7 g piece of palladium, originally at 100.0 °C, placed inside a coffee cup calorimeter containing 115.0 mL of water at 21.8 °C. The specific heat capacity of water is 4.184 J g^-1 °C^-1, the specific heat capacity of palladium is 0.246 J g^-¹ °C^-1, and the density of water is 1.00 g mL^-1.

Question

Quizplus · Accepted Answer

The heat lost by the palladium is equal to the heat gained by the water.$$q_{lost} = q_{gained}$$$$m_{Pd}C_{Pd}(T_f - T_i) = m_{water}C_{water}(T_f - T_i)$$$$(63.7 g)(0.246 J g^{-1} °C^{-1})(T_f - 100.0 °C) = (115.0 g)(4.184 J g^{-1} °C^{-1})(T_f - 21.8 °C)$$$$T_f = 24.3 °C$$

Calculate the Final Temperature of a 63