Chapter 20, Problem 76AP

(a)

To determine

Find the mass of the ice that melts.

Answer to Problem 76AP

The mass of the ice that melts is $15.0\text{ mg}$.

Explanation of Solution

Write the equation for kinetic energy,

    $K_i=\frac{1}{2}m{v}_i^2$                                                                   (I)

Here, $K_i$ is the initial kinetic energy, $m$ is the mass and $v_i$ is the initial velocity.

Write the appropriate energy equation for isolated copper ice system.

    $\begin{array}{l}\Delta K + \Delta E_{\mathrm{int}} = 0 \\ \left(0 - \frac{1}{2}{m}_{\text{Cu}}{v}^2\right) + L_f\Delta m = 0 \\ \Delta m = \frac{m_{\text{Cu}}{v}^2}{\text{2}{L}_f}\end{array}$                                                                   (II)

Here, $\Delta m$ is the change in mass, $m_{Cu}$ is the mass of copper, $v$ is the velocity and $L_f$ is the latent heat of vaporization.

Conclusion:

Substitute $1.60 \text{kg}$ for $m_{Cu}$, $2.50 \text{m/s}$ for $v$ and $3.33\times {10}^5\text{J}/\text{kg}$ for $L_f$ in equation II.

    $\begin{array}{c}\Delta m = \frac{\left(1.60 \text{kg}\right){\left(2.50 \text{m/s}\right)}^2}{\text{2}\left(3.33 \times {10}^5 \text{J/kg}\right)} \\ = 1.50 \times {10}^{-5} \text{kg }\\ \text{=}15.0\text{ mg}\end{array}$

Therefore, the mass of the ice that melts is $15.0\text{ mg}$.

(b)

To determine

Find the energy input, change in internal energy and change in the mechanical energy of the block-ice system.

Answer to Problem 76AP

The general continuity equation for energy is,

    $\Delta K+\Delta E_{\text{int}}=Q$

The energy input, change in internal energy and change in the mechanical energy of the block-ice system are $0$ , $0$ and $-5.00\text{ J}$ respectively.

Explanation of Solution

For the block as a system $Q=0$ such as no energy transfer by heat since there is no temperature difference and $\Delta E_{\mathrm{int}}=0$ such as no temperature or change of state.

Conclusion:

For the block-ice system.

    $\begin{array}{c} \Delta E_{\text{mech}}=-\Delta K \\ =-5.00\text{ J}\end{array}$

Therefore, the energy input, change in internal energy and change in the mechanical energy of the block-ice system are $0$ , $0$ and $-5.00\text{ J}$ respectively.

(c)

To determine

Find the energy input and change in internal energy for the ice system.

Answer to Problem 76AP

The energy input and change in internal energy for the ice system are $0$ and $5.00\text{ J}$ respectively.

Explanation of Solution

For the ice as a system $Q=0$ such as no energy transfer by heat since there is no temperature difference.

Conclusion:

For the ice system.

    $\begin{array}{c} \Delta E_{\text{int}}=\Delta m{L}_f\\ =5.00\text{ J}\end{array}$

Therefore, the energy input and change in internal energy for the ice system are $0$ and $5.00\text{ J}$ respectively.

(d)

To determine

Find the mass of the ice that melts.

Answer to Problem 76AP

The mass of the ice that melts is $15.0\text{ mg}$.

Explanation of Solution

This is same as solved in part a, use the equations in part a.

Conclusion:

Substitute $1.60 \text{kg}$ for $m_{Cu}$, $2.50 \text{m/s}$ for $v$ and $3.33\times {10}^5\text{J}/\text{kg}$ for $L_f$ in equation II.

    $\begin{array}{c}\Delta m = \frac{\left(1.60 \text{kg}\right){\left(2.50 \text{m/s}\right)}^2}{\text{2}\left(3.33 \times {10}^5 \text{J/kg}\right)} \\ = 1.50 \times {10}^{-5} \text{kg }\\ \text{=}15.0\text{ mg}\end{array}$

Therefore, the mass of the ice that melts is $15.0\text{ mg}$.

(e)

To determine

Find the energy input, change in internal energy for the ice system and change in mechanical energy for the block-ice system.

Answer to Problem 76AP

The energy input, change in internal energy for the ice system and change in mechanical energy for the block-ice system are $0$, $5.00\text{ J}$ and $-5.00\text{ J}$ respectively.

Explanation of Solution

For the ice as a system $Q=0$ such as no energy transfer by heat since there is no temperature difference.

Conclusion:

For the ice system.

    $\begin{array}{c} \Delta E_{\text{int}}=\Delta m{L}_f\\ =5.00\text{ J}\end{array}$

For block-ice system

    $\begin{array}{c} \Delta E_{\text{mech}}=-\Delta K \\ =-5.00\text{ J}\end{array}$

Therefore, the energy input, change in internal energy for the ice system and change in mechanical energy for the block-ice system are $0$, $5.00\text{ J}$ and $-5.00\text{ J}$ respectively.

(f)

To determine

Find the energy input and change in internal energy for the metal-sheet system.

Answer to Problem 76AP

The energy input and change in internal energy for the metal-sheet system are $0$ and $0$ respectively.

Explanation of Solution

For the ice as a system $Q=0$ such as no energy transfer by heat since there is no temperature difference.

Conclusion:

For the metal sheet system, $\Delta E_{\text{int}}=0$ such as no change in state or temperature.

Therefore, the energy input and change in internal energy for the metal-sheet system are $0$ and $0$ respectively.

(g)

To determine

Find the change in the temperature.

Answer to Problem 76AP

The change in the temperature is $4.04 \times {10}^{-3}\text{ °C}$.

Explanation of Solution

Write the appropriate energy equation for copper-copper system.

    $\Delta K + \Delta E_{\mathrm{int}} = 0 \to \Delta E_{\mathrm{int}} = -\Delta K$                                                                  (III)

As the system have symmetry, each of the copper slab possesses half of the internal energy change of the system.

    $\Delta E_{\mathrm{int},\text{copper}} = \frac{1}{2}\Delta E_{\mathrm{int}} = -\frac{1}{2}\Delta K = -\frac{1}{2}\left(0 - \frac{1}{2}m{v}^2\right) =\frac{1}{4}m{v}^2$                                     (IV)

Then, the internal energy change of the copper slab is,

    $\begin{array}{l}\Delta E_{\mathrm{int},\text{copper}}=mc\Delta T=\frac{1}{4}m{v}^2\\ \Delta T=\frac{v^2}{4c}\end{array}$                                      (V)

Conclusion:

Substitute $2.50 \text{m/s}$ for $v$ and $387 \text{J/kg }\cdot \text{ °C}$ for $c$ in equation V.

    $\begin{array}{c}\Delta T = \frac{{\left(2.50 \text{m/s}\right)}^2}{\text{4}\left(387 \text{J/kg }\cdot \text{ °C}\right)}\\ = 4.04 \times {10}^{-3}\text{ °C}\end{array}$

Therefore, the change in the temperature is $4.04 \times {10}^{-3}\text{ °C}$.

(h)

To determine

Find the energy input, change in internal energy for the sliding slab and change in mechanical energy for the two-slab system.

Answer to Problem 76AP

The energy input, change in internal energy for the sliding slab and change in mechanical energy for the two-slab system are $0$, $2.50\text{ J}$ and $-5.00\text{ J}$ respectively.

Explanation of Solution

For the sliding slab $Q=0$ such as no energy transfer by heat since there is no temperature difference and $\Delta E_{\mathrm{int}}=2.50\text{ J}$ such as the friction transfer kinetic energy into internal energy.

Conclusion:

For two-slab system

    $\begin{array}{c} \Delta E_{\text{mech}}=\Delta K \\ =-5.00\text{ J}\end{array}$

Therefore, the energy input, change in internal energy for the sliding slab and change in mechanical energy for the two-slab system are $0$, $2.50\text{ J}$ and $-5.00\text{ J}$ respectively.

(i)

To determine

Find the energy input and change in internal energy for the stationary slab.

Answer to Problem 76AP

The energy input and change in internal energy for the stationary slab are $0$ and $2.50\text{ J}$ respectively.

Explanation of Solution

For the stationary slab $Q=0$ such as no energy transfer by heat since there is no temperature difference.

Conclusion:

For stationary slab

    $\Delta E_{\text{int}}=2.50\text{ J}$

Therefore, the energy input, change in internal energy for the sliding slab and change in mechanical energy for the two-slab system are $0$, $2.50\text{ J}$ and $-5.00\text{ J}$ respectively.