Asked by AE
An 8.56 kg block of ice at 0 degrees celsius is sliding on a rough horizontal icehouse floor (also at 0 degrees celsius) at 15.6 m/s. Assume that half of any heat generated goes into the floor and the rest goes into the ice. How much ice (in kg) has melted after the speed of the ice has been reduced to 10.2 m/s? What is the maximum amount of ice that will melt?
Answers
Answered by
MathMate
Use energy considerations.
m=8.56 kg
v0=15.6 m/s,
v1=10.2 m/s
latent heat of fusion of ice
= 333.55 KJ/kg
Initial kinetic energy
=(1/2)mv0²
Final kinetic energy
=(1/2)mv1²
Half of energy lost goes to melt the ice block
Ef=(1/2)m(v0²-v1²)
=(1/2)8.56(15.6²-10.2²)
= 596.3 J
Amount of melted ice
= ((596.3/2)/1000)KJ /333.55 (KJ/kg)
= 0.9 g
Maximum amount of ice would melt if all the kinetic loss goes to melting the ice.
m=8.56 kg
v0=15.6 m/s,
v1=10.2 m/s
latent heat of fusion of ice
= 333.55 KJ/kg
Initial kinetic energy
=(1/2)mv0²
Final kinetic energy
=(1/2)mv1²
Half of energy lost goes to melt the ice block
Ef=(1/2)m(v0²-v1²)
=(1/2)8.56(15.6²-10.2²)
= 596.3 J
Amount of melted ice
= ((596.3/2)/1000)KJ /333.55 (KJ/kg)
= 0.9 g
Maximum amount of ice would melt if all the kinetic loss goes to melting the ice.
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