In an experiment, 20.0 g of ice at 0.00 C is quickly transferred to a calorimeter at 24.20 C.The final temperature comes to 2.60 C.What is the approximate heat capacity of the calorimeter? [ $\Delta$ H_fus = 334 J/g; c_pice) = 2.06 J/(g . C); c_p(water) = 4.18 J/(g . C)]

Question

Quizplus · Accepted Answer

The heat absorbed by the ice to reach 0°C, melt, and then warm up to 2.6°C is equal to the heat lost by the calorimeter. The heat capacity of the calorimeter can be calculated using the formula $q = mc\Delta T$ for the ice and water phases separately, and then setting this equal to $C\Delta T$ for the calorimeter, where $C$ is its heat capacity. The correct calculation leads to a heat capacity of approximately 319 J/°C for the calorimeter.

In an Experiment, 20 Δ\DeltaΔ Hfus = 334 J/g; Cpice) = 2

In an Experiment, 20 $\Delta$ H_fus = 334 J/g; C_pice) = 2