Math, asked by chinkalsankhla, 8 months ago

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

Answered by bangarusamhi
1

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."

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