Solution.
l=0.8m;
A=4.0⋅10−3m2;
ΔT=100oC=100K;
tm=5.5⋅10−4kgs−1;
K=401Js−1m−1K−1;
Lf−?;
ΔtΔQ=(tm)Lf;
ΔtΔQ=lΔT⋅K⋅A ;
ΔtΔQ=0.8m100K⋅401Js−1m−1K−1⋅4.0⋅10−3m2=200.5Js−1;
Lf=tmΔtΔQ ;
Lf=5.5⋅10−4kgs−1200.5Js−1=3.65⋅105Jkg−1;
Answer: Lf=3.65⋅105Jkg−1.
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