That represents an additional reduction of the water temperature that wasn't in those examples. You might think this temperature is low in light of our previous examples, but recall that in melting, the ice extracts an extra 333 J/g of energy from the water without a change it its temperature. Now adding all of these steps and setting the result equal to zero, according to the law of conservation of energy, will get us to our final temperature. They meet at the equilibrium temperature, T f. The red line shows the three parts of the heating process for the ice, and the blue one shows the change in the water temperature. If the two masses were the same, we'd expect (because the specific heats are the same) that the equilibrium temperature would be the average of 80˚C and 19˚C, 49.5˚C. Is it reasonable for 80 grams of hot water to only raise the temperature of 500 grams of 19˚C water by only ten degrees? That actually seems about right because the larger mass of water is 5× the smaller one. One last thing here: We should always ask whether our solution makes sense. The first law of thermodynamics is introduced and described. $$|q_
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