To one liter of water at an initial temperature of 20℃ is added 200 gm of ice at -5℃.

To one liter of water at an initial temperature of 20℃ is added 200 gm of ice at -5℃. Determine the final equilibrium temperature of the mixture, assuming the water/ice system is thermally isolated from its environment.

Expert Solution

The warmer water must provide all the energy for this to happen. Keeping in mind that the heat lost by the warmer must equal the heat gained by the colder (q_lost = q_gain), we have this:

heat to warm ice 10 degrees + heat to melt ice + heat to warm cold water by unknown amount = heat lost by the warm water

Here are the numbers:

[(200)(5)(2.06)] + [(6020)(200.0/18.0)] + [(200)(x - 0)(4.184)] = (1000.0)(20 - x)(4.184)

2060 +66888+836.8x = 83680 - 4184x

Notice I used 6020 J rather than 6.02 kJ for the molar heat of fusion. That is because the other two parts of the left-hand side of the equation will give Joules as their answer. I used 6020 so that all three parts would be in Joules. If I had used 6.02, then that middle part woud have been in units of kJ.

Notice the use of x - 0 and 20 - x. This time, visualize (or write out) the number line used above. The zero is to the left, the 60.0 to the right and the x is in between the 0 and the 60.0.

Proceeding with the solution, I get:

2060 +66888+836.8x = 83680 - 4184x

14732 = 3347.2 x

and so, x = 4.40

genius_generous answered 2 years ago

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