Question

A 2.55 g lead bullet at 21.2 ◦C is fired at a speed of 212 m/s into a large block of ice at 0 ◦C, into which it embeds itself. What quantity of ice melts? Assume the specific heat of lead is 128 J/kg · ◦C and the latent heat of fusion of water 3.33 × 105 J/kg. Answer in units of g.

Answer 1

Answer:

0.192 g of ice will melt.

Explanation:

Applying the law of conservation of energy to the given situation, we get:

Heat Energy lost by Bullet + Kinetic Energy of Bullet = Heat Gained by Ice

m₁CΔT + (1/2)m₁v² = m₂L

where,

m₁ = mass of bullet = 2.55 g = 2.55 x 10⁻³ kg

C = specific heat of lead = 128 J/kg

ΔT = Difference in temperature of bullet and ice = 21.2°C

v = speed of bullet = 212 m/s

L = latent heat of fusion o water = 3.33 x 10⁵ J/kg °C

Therefore,

(2.55 x 10⁻³ kg)(128 J/kg °C)(21.2°C) + (1/2)(2.55 x 10⁻³ kg)(212 m/s)² = m₂(3.33 x 10⁵ J/kg)

m₂ = 64.22 J/3.33 x 10⁵ J/kg

m₂ = 1.92 x 10⁻⁴ kg = 0.192 g

Answer 2

Answer:

$m_{ice} = 192.863\,g$

Explanation:

The bullet is stopped by the ice block, which cools down the first one. The energy is dissipated in the form of heat, which melts part of the ice block. This phenomenon is modelled after the First Law of Thermodynamic, which is a generalization of the Principle of Energy Conservation:

$-Q_{out} + K_{in} - K_{out} + U_{in} - U_{out} = 0$

$Q_{out} = K_{in} - K_{out} + U_{in} - U_{out}$

$m_{ice}\cdot L_{f} = \frac{1}{2}\cdot m_{bullet} \cdot (v_{o}^{2}-v_{f}^{2}) + m_{bullet}\cdot c\cdot (T_{o} - T_{f})$

The mass of ice that is melt is:

$m_{ice}= \frac{m_{bullet}\cdot \left[\frac{1}{2}\cdot (v_{o}^{2}-v_{f}^{2}) + c\cdot (T_{o}-T_{f})\right]}{L_{f}}$

$m_{ice} = \frac{(2.55\,g)\cdot \left\{\frac{1}{2}\cdot \left[(212\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]+\left(128\,\frac{J}{kg\cdot ^{\textdegree}C} \right)\cdot (21.2\,^{\textdegree}C - 0\,^{\textdegree}C) \right\}}{3.33\times 10^{2}\,\frac{J}{g} }$

$m_{ice} = 192.863\,g$