Can someone help me with this chemistry problem?
61.20g of water at 58.0 degrees C is added to a student calorimeter containing 48.80g of water at 25.0 degrees C. What is the final temperature of the water mixture?
Best Answer - Chosen by Asker
When the water is mixed, the heat energy lost by the hotter water equals to the heat energy gain by the colder water.
Let the final temperature of the water mixture be t.
Heat lost = 61.20 x c x (58 - t)
Heat gain = 48.8 x c x (t - 25)
61.20 x c x (58 - t) = 48.8 x c x (t - 25)
3549.6 - 61.20t = 48.80t - 1220
110t = 4769.6
t = 43.36 degree C
Let the final temperature of the water mixture be t.
Heat lost = 61.20 x c x (58 - t)
Heat gain = 48.8 x c x (t - 25)
61.20 x c x (58 - t) = 48.8 x c x (t - 25)
3549.6 - 61.20t = 48.80t - 1220
110t = 4769.6
t = 43.36 degree C
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