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SIGH... physics help once again, please!!

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My son is studying Temperature, heat, and expansion, heat transfer,change of phase, and thermodynamics in physics. Here are some of the problems he's having trouble with:


1. A typical woman gives off heat at the rate of about 8000 calories/hour. How long would a woman have to stay in a bath of 60,000g of 26 degree C water in order to raise the water temperature to 30 degrees C? Presume that all the heat given off by the woman is transferred to the water, and that the water does not lose any heat to the air. show calculations.


We think the formula is Q=mc* change in temperature.

the specific heat of water is equal to 1calorie/g degree C.


2. The specific heat of mercury is .03 calories/g degree C and its boiling point is 357 degrees C. he specific heat of water is 1 calorie/g degreeC. It takes 65 calories of energy to vaporize one gram of mercury and 540 calories to vaporize 1 gram of water. If both substances begin at room temperature (about 22 degrees C), does it take more energy to boil a gram of mercury or a gram of water? show calculations.


3. In 1783 the Montgolfier brothers of France launched what is possibly the first balloon flight carrying passengers--- a duck, a rooster, and a sheep. Their balloon about 35 feet in diameter and constructed of cloth lined with paper, was launched by filling it with smoke. The flight landed safely some 8 minutes later. Explain the physics of this flight (that is, its ascent, descent, and landing.


4. In sub-tropical climates, temperatures occasionally drop below 0 degree C at night. Gardeners in these regions sometimes try to water their plants at night. just before the chill sets in. Why would increasing the water content of a plant make it more resistant to drops in temperature?


I know it's a lot to ask, but we're really stumped and any help would be greatly appreciated.:001_smile:

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For Problem #1:


The heat required to raise the bath water from 22 to 26 deg. C is


(1 cal/g/degC)(60,000 g)(26-22 deg) = 240,000 cal


240,000 cal/ 8,000 cal/hr = 30 hr


For Problem # 2:


The heat to vaporize 1 g of Hg is (0.03 cal/g/deg C)(357-22 deg C)(1 g) + (65 cal/g)( 1 g) = 75.05 cal


The heat to vaporize 1 g of H20 is (1 cal/g/deg C)(100-22 deg C)(1 g) + (540 cal/g)(1 g) = 618 cal


For Problem # 4:


I think the answer has somethiing to do with the fact that water has such a high heat capacity, that adding water to the plant means that more energy is required to lower the plant temperature...


Let me know if you need any more explanation -- or, heaven forbid, if these aren't the right answers!

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My son and I are very much appreciative for your help. We attempted question 1, but were unsure of the answer. You confirmed that we were right!


Thanks again, Martha!!

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