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_p(ice)= 2.06 J/(g.°C); c_p(water)= 4.18 J/(g.°C)]

Question

Quizplus · Accepted Answer

First, we need to determine the amount of heat that is required to melt the ice and then to warm it up to the final temperature. The calculation is as follows:

q = (20.0 g) x (334 J/g) + (20.0 g) x (2.06 J/(g°C)) x (2.60°C) = 14,602 J

Here, we used the heat of fusion of ice (334 J/g) and the specific heat capacity of ice (2.06 J/(g°C)).

Next, we can use the fact that the amount of heat lost by the ice is equal to the amount of heat gained by the calorimeter and the water. We can write:

q = -(C x ΔT)

where C is the heat capacity of the calorimeter and ΔT is the change in temperature of the calorimeter and the water.

Rearranging, we get:

C = -q/ΔT = -14,602 J / (24.20°C - 2.60°C) = 318.96 J/°C

Therefore, the approximate heat capacity of the calorimeter is 319 J/°C, which is closest to choice A.

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

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