Q1) You cool a 130.0 g slug of red-hot iron (temperature 745 ? C ) by dropping i

Question

Q1) You cool a 130.0g slug of red-hot iron (temperature 745 ?C) by dropping it into an insulated cup of negligible mass containing 70.0g of water at 20.0 ?C. Assume no heat exchange with the surroundings.

Part A

What is the final temperature of the water?

Express your answer using three significant figures.

Part B

What is the final mass of the iron and the remaining water?

Express your answer using three significant figures.

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Q2) Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0.200W/(m?K) ] surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere 1.60m in diameter having a layer of fat 4.10cm thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at 2.60?C during hibernation, while the inner surface of the fat layer is at 31.2?C.

Assume the surface area of each layer is constant and given by the surface area of the spherical model constructed for the black bear.

Part A

What should the temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 52.0W ?

Part B

How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 52.0W ?

Explanation / Answer

2nd

First, you need to get the square meters of the surface of the idealized "bear-sphere".

Second, you need to convert the fat layer thickness to meters.

Third, Heat rate = thermal conductivity * area * /T / thickness, solve for delta-T, and you know that the high temp side is 31.2

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