- #1
user0n9
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an IRON cube with (Length 1.3cm Width 1.3cm Height 0.7cm) so
Volume = 1.183cm^3
density of an IRON = 7.8gm/cm^3 so
Mass = 9.2274gram
Q=mcT c for iron = 0.450 J/(g·K), T difference = 160 degree celsius
Q=9.2 . 0.45 . 160
Q=662.4Joule
1Joule = 1 Watt.second
so if i need to heat up an IRON from 20C to 180C (T difference = 160 degree celsius) in 60second
so i need (662.4/60) = 11.04Watt
correct? please advice..
thanks
Volume = 1.183cm^3
density of an IRON = 7.8gm/cm^3 so
Mass = 9.2274gram
Q=mcT c for iron = 0.450 J/(g·K), T difference = 160 degree celsius
Q=9.2 . 0.45 . 160
Q=662.4Joule
1Joule = 1 Watt.second
so if i need to heat up an IRON from 20C to 180C (T difference = 160 degree celsius) in 60second
so i need (662.4/60) = 11.04Watt
correct? please advice..
thanks