An insulated container s used to hold 40.10 g of water at 36.20 degree C. A samp

Question

An insulated container s used to hold 40.10 g of water at 36.20 degree C. A sample of copper weighing 13.40 g is placed In a dry test tube and heated for 30 minutes in a boiling water bath at 100.10C. Th. healed test tube Is carefully removed from the water bath with laboratory tongs and Inclined so that the copper slides into the water in the Insulated container. Given that the specific heat or solid copper is 0.385 J/(g.degree C). calculate the maximum temperature of the water in the insulated container after the copper metal is added.

Explanation / Answer

heat lost = heat gain

mCpdT(metal) = mCpdT(water)

13.40 x 0.385 x (100.10 - Tf) = 40.10 x 4.184 x (Tf - 36.20)

516.416 - 5.16Tf = 167.78Tf - 6073.58

6590 = 172.94Tf

Tf = 38.106 oC

So the maximum temperature or the final temperature of water would be 38.106 oC

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