The Bullet and the Ice

Question

A 3.29g aluminum bullet at 31.5°C is fired at a speed of 229 m/s into a large block of ice at 0°C, in which it embeds itself. What quantity of ice melts?

Answer 1

We will be using the law of conservation of energy. The energy released by bullet will be equal to energy that required to melt ice.

The bullet was moving in the beginning; as a result it had kinetic energy K = ½( m₁ v²),

where m₁ = 3.29 x 10^-3 kg – mass of the bullet,

v = 229 m/s –  its speed.

The temperature of the bullet is t₁ = 31. 5 °C. The melting point of ice is t₂ = 0 °C. Hence, while cooling to this temperature it will give the ice extra heat Q₁ = c∙m₁∙( t₁ – t₂),

here c = 900 J/(kg∙°C) - specific heat of aluminum.

The total amount of energy received from the bullet by ice is K + Q₁. This energy will be used to melt some ice. Ice required heat Q₂ = L∙m₂ to melt, where

L = 3.34 x 10 ⁵ J/kg – heat of fusion of ice.

Now we can write

                                           ½( m₁ v²) +  c∙m₁∙( t₁ – t₂)  = L∙m₂ .

From this equation

                                              m₂ = ( (½( m₁v²) +  c∙m₁∙( t₁ – t₂) )  / L   

        

Now you need to put all values of variables in this equation and get the answer.

If my calculator is not wrong, the answer is 5.3 x 10 ^-,4 kg or 0.53 g.