Feel free to answer only one, but this is my last question ofthe day so I figure

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Feel free to answer only one, but this is my last question ofthe day so I figured I'd throw in both of these just in casesomeone was nice enough to help with both. I don'tnecessarily need a thorough explanation...
The latent heat of vaporizaiton of H2O at bodytemperature (37.0°C) is 2.42 x 106 J/kg. Tocool the body of a 79 kg jogger(average specific heat capacity = 3500 J/(kg·C°) by1.4 C°, how many kilograms ofwater in the form of sweat have to be evaporated? A 0.235 kg piece of aluminum thathas a temperature of -155°C is added to 1.5 kg of water thathas a temperature of 1.5°C. Atequilibrium the temperature is 0.0°C. Ignoring the containerand assuming that the heat exchanged with the surroundings isnegligible, determine the mass of water that has been frozen intoice. (See Table 12.2and Table 12.3 forappropriate constants.)
Feel free to answer only one, but this is my last question ofthe day so I figured I'd throw in both of these just in casesomeone was nice enough to help with both. I don'tnecessarily need a thorough explanation...
The latent heat of vaporizaiton of H2O at bodytemperature (37.0°C) is 2.42 x 106 J/kg. Tocool the body of a 79 kg jogger(average specific heat capacity = 3500 J/(kg·C°) by1.4 C°, how many kilograms ofwater in the form of sweat have to be evaporated? A 0.235 kg piece of aluminum thathas a temperature of -155°C is added to 1.5 kg of water thathas a temperature of 1.5°C. Atequilibrium the temperature is 0.0°C. Ignoring the containerand assuming that the heat exchanged with the surroundings isnegligible, determine the mass of water that has been frozen intoice. (See Table 12.2and Table 12.3 forappropriate constants.)

Explanation / Answer

Qjogger + QH20 = 0 mjoggercjoggerT +mwatercwaterT +Lwatermwater = 0 mwater = - mjoggercjoggerT /(cwaterT + Lwater) = 79*3500*1.4 /(4180*1.4 + 2.42e6) = ...
the same concept applies to the second question, youll have tofind mice it should be less than 1.5 kg

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