Warm-up 5/8. How much energy is needed to melt exactly 1000 aluminum cans, each with a mass of 14.0 g for recycling. Assume an initial temperature of 26.4 degrees Celsius. (L f = 3.97 x 10 5 J/kg; c p = 8.99 x 10 2 J/kg˚C; MP= 660.4 ˚C). Law of Heat Exchange Q lost = Q gained.
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How much energy is needed to melt exactly 1000 aluminum cans, each with a mass of 14.0 g for recycling. Assume an initial temperature of 26.4 degrees Celsius.
(Lf = 3.97 x 105J/kg; cp= 8.99 x 102 J/kg˚C; MP= 660.4 ˚C)
Note that room temperature in Celsius is about 20°. Re-arranging the equation to solve for the final temperature gives:
The temperature of the coffee doesn't drop by much because the specific heat of water (or coffee) is so much larger than that of steel. This is too hot to drink, but if you wait, heat will be transferred to the surroundings and the coffee will cool.
c tea = c water = 4186 J/kgoC
L f = 3.33 x 105 J/kg and t final = 32oC
Q lost = Q gained tea loses and water gains only melting the ice
(mcDt)tea = (mLf)ice
m ice= (mcDt)tea
m ice = (.180kg)(4186J/kgoC)(32oC)
= 7.2 x 10-2kg