A bullet of mass 10x10-kg moving with a
speed of 20ms- hits an ice block (0°C) of 990g
kept at rest on a frictionless floor and gets
embedded in it. If ice takes 50% of K.E, the
amount of ice melted (in grams) approximately
is (5=4.2J/cal) (Latent heat of ice = 80cal/g)
Answers
Step-by-step explanation:
The bullet has melted 0.003 g of ice.
Given,
Mass of the bullet = 10 g = 10\quad \times \quad { 10 }^{ -3 }=10g=10×10
−3
Speed of the bullet = 20 m/s
Mass of the ice = 990 g
Energy required to melt ice = 50% of kinetic energy
Total kinetic energy required =\quad \frac { 1 }{ 2 } m{ v }^{ 2 }=
2
1
mv
2
=\quad \frac { 1 }{ 2 } \quad \times \quad 10\quad \times \quad { 10 }^{ -3 }\quad \times \quad { 20 }^{ 2 }=
2
1
×10×10
−3
×20
2
Total kinetic energy = 2 J
50% of total kinetic energy =\quad \frac { 20\quad \times \quad 50 }{ 100 }=
100
20×50
50% of total kinetic energy = 1 J
In this case, the latent heat is amount of heat required to solid (ice) into water.
Latent heat of ice = 334 J/g
Amount of energy required to convert
x\quad =\quad \frac { 1 }{ 334 }x=
334
1
x\quad =\quad 2.9\quad \times \quad { 10 }^{ -3 }\quad =\quad 0.003\quad gx=2.9×10
−3
=0.003g
The bullet has melted 0.003 g of ice."