Physics Question, Please Help! Use the worked example above to help you solve th
ID: 1313831 • Letter: P
Question
Physics Question, Please Help!
Use the worked example above to help you solve this problem. At a party, 5.00 kg of ice at -4.90degreeC is added to a cooler holding 30 liters of water at 20.0degreeC. What is the temperature of the water when it comes to equilibrium? Your response differs from the correct answer by more than 10%. Double check your calculations. Degree C What mass of ice at -11.0degreeC is needed to cool a whale's water tank, holding 1.28 times 103 m3 of water, from 20.0degreeC down to a more comfortable 10.0degreeC?Explanation / Answer
Part 1)
By conservation of energy...
The ice will warm to zero, melt, then raise temp
The water will cool off, so...
mc(delta T) + mH + mc(delta T) = mc(delta T)
5(2090)(4.9) + (5)(3.34 X 105) + (5)(4186)(Tf) = 30(4186)(20 - Tf)
51205 + 1.67 X 106 + 20930Tf = 2511600-125580Tf
790395 = 146510Tf
Tf = 5.39o C
(If this is a little off, please let me know. I had to you values from my book and your book may use slightly different values. A fix can be made with a simple comment using the values from your book)
Part 2)
Same idea as part 1
mc(delta T) + mH + mc(delta T) = mc(delta T)
m(2090)(11) + (m)(3.34 X 105) + m(4186)(10) = (1000)(1.28 X 103)(4186)(10)
22990m + 3.34 X 105m + 41860m = 53580800000
361176m = 53580800000
m = 148350 kg
(Again, if this is off its because I had to use the values from my book for ice, Latent Heat of Fusion, and Water) Comment and a fix can be made.