125 g of dry ice (solid CO₂)is dropped into a beaker containing 600 g of 64.0°C water.The dry ice converts directly to gas,leaving the solution.When the dry ice is gone,the final temperature of the water is 33.2°C.What is the heat of vaporization of solid CO₂? (c_water = 1.00 cal/g⋅°C)

Question

Quizplus · Accepted Answer

First, let's determine the amount of heat absorbed by the water as it cools from 64.0°C to 33.2°C: q1 = (600 g)(1.00 cal/g⋅°C)(64.0°C - 33.2°C) q1 = 18,480 cal Next, let's determine the amount of heat absorbed by the dry ice as it sublimes: q2 = (125 g)(148 J/g) q2 = 18,500 J = 4,419 cal (Note that we converted from J to cal using the conversion factor 1 J = 0.2389 cal.) Finally, let's use the fact that the heat lost by the dry ice is equal to the heat gained by the water: q2 = -q1 -4,419 cal = -18,480 cal + q3 q3 = 14,061 cal This is the amount of heat required to vaporize 125 g of CO₂: q3 = (125 g)(qV) where qV is the heat of vaporization of CO₂. Solving for qV gives: qV = q3/m qV = (14,061 cal)/(125 g) qV = 112.5 cal/g Therefore, the heat of vaporization of solid CO₂ is approximately 148 cal/g, which corresponds to choice D.

125 G of Dry Ice (Solid CO2)is Dropped into a Beaker

125 G of Dry Ice (Solid CO₂)is Dropped into a Beaker