How many kilojoules of heat energy are required to heat all the aluminum (Cpof A
ID: 949104 • Letter: H
Question
How many kilojoules of heat energy are required to heat all the aluminum (Cpof Al = .902J/g•°C) in a roll of aluminumfoil (500.0 g) from room temperature (25.0°C) to the temperature of a hot oven (250.0°C)?
One way to cool down your cup of coffee is to plunge an ice-cold piece of aluminum into it. Suppose you store a20.0 g piece of aluminum (Cpof Al = .902J/g•°C) in the refrigerator at 4.40°C and then drop it into your coffee. The coffee temperature drops from 90.0°C to 55.0°C. How many joules of heat energy did the aluminum block absorb? (Ignore the cooling of the cup)
Explanation / Answer
het required = m*s*DT
= 500*0.902*(250-25)
= 101.475 kj
heat absorbed by Al = heat lost by water
= 20*0.902*(55-4.4)
= 912 joule
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