# Ch205 Lesson 5

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)

Answer:

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

*…show more content…*

Answer:

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.)

Answer:

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.

Answer:

X = 0.02 mol

Y = 0.01 mol

Initial T = 21˚C

Final T = 51˚C

51- 21 = +30˚C

4.18J/g˚C * 0.3 * 30˚C = 3.76J * e. Judging on the basis of your answers so far, what is the enthalpy of the reaction 2X(aq)+Y(l) → X2Y(aq)

Answer:

From the previous part:

4.18 kJ