How many joules are absorbed as the temperature of 15.0 grams of aluminum is heated from 20.0oC to 60.0oC?
Q=mcΔTΔT=60.0−20.0=40.0m=15.0 gcAl=0.900 J/g⋅CQ=15.0×0.900×40.0=540 JQ = mcΔT \\ ΔT = 60.0-20.0 = 40.0 \\ m = 15.0 \;g \\ c_{Al} = 0.900 \;J/g \cdot C \\ Q = 15.0 \times 0.900 \times 40.0 = 540 \;JQ=mcΔTΔT=60.0−20.0=40.0m=15.0gcAl=0.900J/g⋅CQ=15.0×0.900×40.0=540J
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