1 of 2 HEAT TRANSFER PRACTICE PROBLEMS The soup in a bowl is too hot to eat Alth

Question

1 of 2 HEAT TRANSFER PRACTICE PROBLEMS The soup in a bowl is too hot to eat Although there are no ice cubes in the freezer, there are several stainless stee spoons that have been stored in the freezer for several hours. By placing the colld spoons in the hot soup, the soup's temperature is lowered from a temperature of 825"C t0 48.5"C. The mass of the soup is 125 g while the mass of each spoon is 45 g A. Assuming that soup has a specific heat of 1.00 calg "C how much heat is transferred from the soup to the spoons? B. If the nitial temperature of the spoons s-15°C and ther specific heat is 0107 caligC. how many spoons ar needed to cool the soup? 0Hint the heat absorbed by the group of spoorns is the same as the heat released by the soup and their final temp is also 48.5" A brass candlestick with a mass of 592 g has an inital temperature of 980C 5043 calories of heat are removed from the candlestick to lower its temperature to 68"C wais the specific heat of brass? A drinking glass with a mass of 385 g is filled with a hot lquid and the liquid transfers 1681 calories of heat to the glass. If the temperature of the glass increases by 22.5 "C what is the specific heat of the glass? The temperature of the air above coastal areas is greatly influenced by the large specific heat of water. The specff heat of ar between temperature of 40"F and 9OF is about 024 calig"Q. Consider the station where 1000 calories of heatare given up by lkg of water causing its temperature to dop by 10 ythe ar oer the waer increases temperature by 55 what is the mass of the air over the water?

Explanation / Answer

Answers:

1. A)

Given c=1cal/(goC), Ti=82.5oC, Tf=48.5oC then delta T=82.5-48.5=34oC

m=125g,

Amount of heat added or removed or transfered Q=c*m*deltaT

Q=1*125*34=4250 calories of heat transfered from soup to spoons.

B) Initial temperature of spoon is -15 oC, c=0.107cal/(goC),

The heat absorbed is same as heat released and final temperature is 48.5oC. So two spoons to be used to cool the soup from 82.5oC to 48.5oC.

2) Given,

Q=5043cal, Ti=98oC, Tf=6.8oC then delta T=98-6.8=91.2oC

m=592g, Q=c*m*deltaT

c=5043/(592*91.2)=0.0934cal/(goC)

3) Given,

Q=1681cal, delta T=22.5oC

m=385g, Q=c*m*deltaT

c=1681/(385*22.5)=0.194cal/(goC)

4) Given,

for water, delta T1=10oC, m1=18, c1=1cal/(goC) and for air delta T2=5.5oC, m2=?, c2=0.24cal/(goC)

m=385g, Q=c*m*deltaT

c1*m1*deltaT1=c2*m2*deltaT2

1*18*10=0.24*m2*5.5

m2=180/1.32=136.36g

5) Given,

Q=915cal, c1=0.55cal/(goC), Ti=?, Tf=18oC

m=225g, Q=c*m*deltaT

915=0.55*225*(Ti-18)

Ti-18=915/123.75=7.394

Ti=7.39+18=25.4oC

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