Physics: 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.

Atoms are most stable when the octet of the outermost shell is complete therefore the answer is c. 8....

A 2.55 g lead bullet at 21.2 ◦C is fired at a speed - id2552126006348, gabrielgutierrez730

Ответ:

randall10

23.01.2022

Physics

$m_{ice} = 192.863\,g$

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:

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

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

$m_{ice} = 192.863\,g$

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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.

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