# Ch205 Lesson 5

1390 words 6 pages
Assignment Chapter 6
Concept Explorations
6.29. Thermal Interactions
Part 1:
In an insulated container, you mix 200. g of water at 80ºC with 100. g of water at 20ºC. After mixing, the temperature of the water is 60ºC. * a. How much did the temperature of the hot water change? How much did the temperature of the cold water change? Compare the magnitudes (positive values) of these changes.
200g of water at 80°C = hot water
100g of water at 20˚C = cold water
After mixing the temperature is 60˚C (equilibrium T)
The temperature of hot water changed:
60˚C - 80˚C = -20˚C
The temperature of cold water changed:
60˚C - 20˚C = 40˚C
The temperature change hot water to cold water is 20:40.

* b. During the mixing, how did

If the temperature of the solution changes from 21ºC to 31ºC, how much heat does the chemical reaction produce? How does this answer compare with that in part a? (You can assume that this solution is so dilute that it has the same heat capacity as pure water.)
100g solution
T changes from initial 21˚C
Final T = 31˚C q = s * m * ∆T
∆T = tf – ti = 31 – 21 = 10˚C q = 4.18J/g˚C * 100g * (+10˚C) = 4.18 kJ
In part (a) the ∆T required 8.36 kJ to raise the temperature 20˚C and part (b) it required 4.18 kJ to raise the temperature 10˚C. Both a and b had 100g of solution and part b temperature raised half of the ∆T than that of part a. Which is consistent with both solution being water.

* c. If you wanted the temperature of 100. g of this solution to increase from 21ºC to 51ºC, how much heat would you have to add to it? (Try to answer this question without using a formula.)
100g solution
Initial T = 21˚C
Final T = 51˚C
∆H = ?
The answer to the is approximately 12.00 kJ * d. If you had added 0.02 mol of X and 0.01 mol of Y to form the solution in part b, how many moles of X and Y would you need to bring about the temperature change described in part c.