hapter 11 ) 130133 ercise 11.86-Enhanced-with Feedback Constants Periodic Table

Question

hapter 11 ) 130133 ercise 11.86-Enhanced-with Feedback Constants Periodic Table Part A A sample of steam with a mass of 0.503 g at a temperature of 100 C condenses into an insulated container holding 4.50 g of water at 2.0 °C. (For water, Hoap 40.7 kJ/mol and Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture? Express your answer using three significant figures. You may want to reference Pages 437-489) Section 11.6 while completing this problem. Submit Request Answer Provide Feedback Next >

Explanation / Answer

Hovap = 40.7 kJ/mol
           = 40700 J/mol

So, heat lost by steam = (0.503 g) x (40700 J/mol)
                                    = (0.503 g) x (40700 J / 18 g)
                                     = 1137 J

Heat gained by cold water = (4.50 g) x (4.18 J g-1oC-1) x (T)

The heat lost by the steam is gained by the cold water

So,

(4.50 g) x (4.18 J g-1oC-1.) x (T) = 1137 J
or, (18.81 J oC-1.) x (T) = 1137 J
or, T = (1137 J) / (18.81 J oC-1.)
or, T = 60.45 oC

But, T = Tfinal - Tinitial
or, 60.45 oC = Tfinal – 2 oC
or, Tfinal = 60.45 oC +2.0 oC
or, Tfinal = 62.45 oC

Now you have two bodies of water, 0.503 g at 100 oC, and 4.50g at 62.45 oC, so find the final temperature using a weighted average:

[ (0.503 g x 100 oC) + (4.50 g x 62.45 oC) ] / (0.524 g + 4.40 g) = 67.3 oC

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