#2172 Ans: 24 degrees celsius What would the evaporation rate (meters of depth p
ID: 1281806 • Letter: #
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#2172
Ans: 24 degrees celsius
What would the evaporation rate (meters of depth per year) be if all of the solar power incident normally (324 cal/sec-m2 ...from problem 2075) went into evaporating water? What does this give for the overall efficiency of the HHE process? (You will have to make some reasonable assumptions to find the average number of seconds of normally incident sunlight over a year. To simplify the assumptions, take the average daily angle of incidence to be 45 degree, the average yearly angle to be the average between maximum angle (on Dec. 21 ...Winter Solstice) and the minimum angle (on June 21); the latitude of Qattara is 30 degree N.; the ecliptic angle (23 degree) cancels. Average cloudiness at Qattara over the year is about 15% of daylight hours). A calorimeter contains 500 grams of water and 300 gm of ice, all at a temperature of 0 degree C. A block of metal of mass 1000 gm is taken from a furnace where its temperature was 240 degree C and is dropped quickly into the calorimeter. As a result, all of the ice is just melted. What would the final temperature of the system have been if the mass of the block had been twice as great? Neglect heat loss from the calorimeter, and the heat capacity of the calorimeter.Explanation / Answer
The change in system is conversion of ice to water and reduction of temperature of metal block to 0 degrees. Writing the heat equation for given system,
0.3*334*1000=0.1*240*C
C=4175J/kgK
Now if mass of metal block= 0.2
heat equation is now written as,
0.2*4175*(513-t)=0.3*334*1000+ 0.8*4184*(t-273)
t=(428355+913785-100200)/4182.2= 296.9K=23.96degreeC=24degreeC
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