48 g of ice at -10oC is added to 150 g glass filled with 250 g of water at 20oC.

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Explanation / Answer

Q1 = 0oC -> 10oC (ice) = (mcT)ice = (48g)(0.5 cal/goC)(10oC) = 240 cal Q2 = melt ice = m x Lf = (48g)(79.6 cal/g) = 3821 cal This number adds up to 4061 cal. Now we compare this number to Q3: Q3 = 20oC -> 0oC (warm water and glass vessel) = (mcT) warm water + (mcT)glass vessel = (250 g)(1 cal/goC)(20oC) + (150g)(0.2cal/goC)(20oC) = 5600 cal = 5000 cal Since 4061 cal is less than 5600 cal, we know that the final temperature is between 0 degrees C and 20 degrees C. Also, the final equilibrium consists of all liquid water. Now we find the final temperature: Qlost = Qgained (mcT)warm water + (mcT)glass vessel = (mcT)cool water + Q1 + Q2 (250g)(1cal/goC)(20oC-Tf) + (150g)(0.2cal/goC)(20oC-Tf) = (48g)(1cal/goC)(Tf-0oC) + 4061 cal Now we solve for Tf and end up with 4.69 degrees C.

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