An 8.54kg block of ice at 0 ∘C is sliding on a rough horizontal icehouse floor (also at 0 ∘C) at 15.2m/s .?
Assume that half of any heat generated goes into the floor and the rest goes into the ice.
How much ice melts after the speed of the ice has been reduced to 10.3m/s ?
m = _____________kg
What is the maximum amount of ice that will melt?
m=______________kg
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Verified answer
Q=ΔK
ΔK=1/2m(vi^2-vf^2)
ΔK=0.5*8.54(12.5^2-10.3^2)=214.2 J
the heat to melt the ice is
Q=mλ
λ=3.3x10^5 J/kg
then
m=(ΔK/2)/λ
m=(214.2/2)/3.3x10^5=3.25 x10^-4 kg =0.325 g
the maximum amount of ice that will melt is when vf=0
then
m=λ/(ΔKmax/2)
ΔKmax=1/2mvi^2=985.5
m=(985.5/2)3.3x10^5=1.49x10^-3 kg = 1.49 g