Question

15. A coffee cup calorimeter contains 400.0 g of water and 25.0 g of ice at 0.0 °C. When a 35.00 g scoop of calcium chloride (also at 0.0 °C) is added to the ice water, the calcium chloride dis- solves as shown. The heat produced by this reaction is enough to melt all of the ice and warm up the resulting calcium chloride solution. What is the final temperature of this solution? Assume there is no heat lost to the surroundings. The specific heat of the calcium chloride solution is 4.2 J/gºC. Other potentially useful constants are given below. CaCl (5) - Cat(aq) + 2 CH(aq) AH = -81 kJ specific heat of steam = 2.01 J/gºC specific heat of water = 4.18 J/g °C specific heat of ice = 2.09 J/gºC heat of fusion of H,O=334 J/g heat of vaporization of H,0 = 2260 Jig

Answer 1

35.00 grams of calcium chloride (molar mass 111 g/mol) contains $\frac{35.00 \ g}{111 \ g/mol}=0.315 \ mol$

Heat released by dissolution of 0.315 moles of calcium chloride $=0.315 \ mol \times 81 \ kJ/mol \times 1000 \ J/kJ=25541 \ J$

Heat required to melt 25.0 grams of ice at 0.0 deg C

25.5 g x 334 J/g = 8517 J

Heat released by dissolution of 0.315 moles of calcium chloride minus heat required to melt 25.0 grams of ice at 0.0 deg C

Total amount of calcium chloride solution includes sum of masses of calcium chloride, water and ice.

Total amount of calcium chloride solution

Let T deg C be the final temperature of calcium chloride solution

$q=W \times S \times \Delta T$

$17024 \ J=460 \ J \times 4.2 \ \frac{J}{g ^oC} \times \left ( T-0.0 \right ) \ ^oC$

$\left ( T-0.0 \right ) \ ^oC=8.8 \ ^oC$