Chapter 20, Problem 20.76AP

(a)

To determine

The mass of the ice that melts.

Answer to Problem 20.76AP

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

Explanation of Solution

Given info: The mass of the copper block is $1.60\text{kg}$, the initial velocity of the copper block is $2.50\text{m}/\text{s}$, the final velocity of the block is $0\text{m}/\text{s}$, the temperature of the ice is $0°\text{C}$, the temperature of the copper block is $0°\text{C}$ and the temperature of the ice is $0°\text{C}$.

Write the expression for the change in kinetic energy of the block.

$\Delta K=\frac{1}{2}m\left(v_{\text{f}}^2-v_{\text{i}}^2\right)$

Here,

$v_{\text{f}}$ is the final velocity of the block.

$v_{\text{i}}$ is the initial velocity of the block.

Substitute $0\text{m}/\text{s}$ for $v_{\text{f}}$, $2.50\text{m}/\text{s}$ for $v_{\text{i}}$ and $1.60\text{kg}$ for $m$ in the above equation.

$\begin{array}{c}\Delta K=\frac{1}{2}\left(1.60\text{kg}\right)\left({\left(0\text{m}/\text{s}\right)}^2-{\left(2.50\text{m}/\text{s}\right)}^2\right)\\ =-5\text{J}\end{array}$

Thus, the change in kinetic energy of the block is $-5\text{J}$.

Write the expression for the change in internal energy.

$\Delta E_{\text{int}}=L\Delta m$ (1)

Here,

$L$ is the latent heat of the fusion of ice.

$\Delta m$ is the mass of ice that melts.

Substitute $3.33\times {10}^5\text{J}/\text{kg}$ for $L$ in the above equation.

$\Delta E_{\text{int}}=\left(3.33\times {10}^5\text{J}/\text{kg}\right)\Delta m$

Write the expression for the conservation of energy for the isolated copper ice system.

$\Delta K+\Delta E_{\text{int}}=0$

Here,

$\Delta K$ is the change in the kinetic energy.

$\Delta E_{\text{int}}$ is the change in the internal energy.

Substitute $\left(3.33\times {10}^5\text{J}/\text{kg}\right)\Delta m$ for $\Delta E_{\text{int}}$ and $-5\text{J}$ for $\Delta K$ in above equation.

$\begin{array}{c}-5\text{J}+\left(3.33\times {10}^5\text{J}/\text{kg}\right)\Delta m=0\\ \Delta m\approx 1.50\times {10}^{-5}\text{kg}\times \frac{{10}^6\text{mg}}{1\text{kg}}\\ =15.0\text{mg}\end{array}$

Conclusion:

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

(b)

To determine

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

Answer to Problem 20.76AP

The input energy is $0$, the change in internal energy is $0$ and the change in mechanical energy for the block-ice system is $5\text{J}$.

Explanation of Solution

Given info: The mass of the copper block is $1.60\text{kg}$, the initial velocity of the copper block is $2.50\text{m}/\text{s}$, the final velocity of the block is $0\text{m}/\text{s}$, the temperature of the air is $0°\text{C}$, the temperature of the copper block is $0°\text{C}$ and the temperature of the ice is $0°\text{C}$.

The temperature of the air is $0°\text{C}$, the temperature of the copper block is $0°\text{C}$ and the temperature of the ice is $0°\text{C}$. There is no energy transfer by heat because there is no temperature difference between the block and the ice. The change in internal energy is $0$ because the state of the material is not changed due to the same temperature.

Write the expression for the change in mechanical energy of the block.

$\Delta E_{\text{mechanical}}=-\Delta K$

From part (a), change in kinetic energy of the block is $-5\text{J}$.

Substitute $-5\text{J}$ for $\Delta K$ in the above equation.

$\begin{array}{c}\Delta E_{\text{mechanical}}=-\left(-5\text{J}\right)\\ =5\text{J}\end{array}$

Conclusion:

Therefore, the input energy is $0$, the change in internal energy is $0$ and the change in mechanical energy for the block-ice system is $5\text{J}$.

(c)

To determine

The input energy and the change in internal energy for the ice as a system.

Answer to Problem 20.76AP

The input energy is $0$ and the change in internal energy for the ice as a system is $5\text{J}$.

Explanation of Solution

Given info: The mass of the copper block is $1.60\text{kg}$, the initial velocity of the copper block is $2.50\text{m}/\text{s}$, the final velocity of the block is $0\text{m}/\text{s}$, the temperature of the air is $0°\text{C}$, the temperature of the copper block is $0°\text{C}$ and the temperature of the ice is $0°\text{C}$.

The temperature of the ice is $0°\text{C}$. The temperature is constant for ice that means there is no temperature difference. There is no energy transfer by heat because there is no temperature difference for ice. But the state of the ice is changed when it melts, so the internal energy is not $0$.

From equation (1),  the expression for the change in internal energy.

$\Delta E_{\text{int}}=L\Delta m$

From part (a), the mass of the ice that melts is $15.0\text{mg}$.

Substitute $3.33\times {10}^5\text{J}/\text{kg}$ for $L$ and $15.0\text{mg}$ for $\Delta m$ in the above equation.

$\begin{array}{c}\Delta E_{\text{int}}=\left(3.33\times {10}^5\text{J}/\text{kg}\right)\left(15.0\text{mg}\times \frac{{10}^{-6}\text{kg}}{1\text{mg}}\right)\\ \approx 5\text{J}\end{array}$

Conclusion:

Therefore, the input energy is $0$ and the change in internal energy for the ice as a system is $5\text{J}$.

(d)

To determine

The mass of the ice that melts.

Answer to Problem 20.76AP

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

Explanation of Solution

Given info: The mass of the copper block is $1.60\text{kg}$, the initial velocity of the copper block is $2.50\text{m}/\text{s}$, the final velocity of the block is $0\text{m}/\text{s}$, the temperature of the ice is $0°\text{C}$, the temperature of the copper block is $0°\text{C}$ and the temperature of the ice is $0°\text{C}$.

Write the expression for the change in kinetic energy of the block.

$\Delta K=\frac{1}{2}m\left(v_{\text{f}}^2-v_{\text{i}}^2\right)$

Here,

$v_{\text{f}}$ is the final velocity of the block.

$v_{\text{i}}$ is the initial velocity of the block.

Substitute $0\text{m}/\text{s}$ for $v_{\text{f}}$, $2.50\text{m}/\text{s}$ for $v_{\text{i}}$ and $1.60\text{kg}$ for $m$ in the above equation.

$\begin{array}{c}\Delta K=\frac{1}{2}\left(1.60\text{kg}\right)\left({\left(0\text{m}/\text{s}\right)}^2-{\left(2.50\text{m}/\text{s}\right)}^2\right)\\ =-5\text{J}\end{array}$

Thus, the change in kinetic energy of the block is $-5\text{J}$.

Write the expression for the change in internal energy.

$\Delta E_{\text{int}}=L\Delta m$ (1)

Here,

$L$ is the latent heat of the fusion of ice.

$\Delta m$ is the mass of ice that melts.

Substitute $3.33\times {10}^5\text{J}/\text{kg}$ for $L$ in the above equation.

$\Delta E_{\text{int}}=\left(3.33\times {10}^5\text{J}/\text{kg}\right)\Delta m$

Write the expression for the conservation of energy for the isolated copper ice system.

$\Delta K+\Delta E_{\text{int}}=0$

Here,

$\Delta K$ is the change in the kinetic energy.

$\Delta E_{\text{int}}$ is the change in the internal energy.

Substitute $\left(3.33\times {10}^5\text{J}/\text{kg}\right)\Delta m$ for $\Delta E_{\text{int}}$ and $-5\text{J}$ for $\Delta K$ in above equation.

$\begin{array}{c}-5\text{J}+\left(3.33\times {10}^5\text{J}/\text{kg}\right)\Delta m=0\\ \Delta m\approx 1.50\times {10}^{-5}\text{kg}\times \frac{{10}^6\text{mg}}{1\text{kg}}\\ =15.0\text{mg}\end{array}$

Conclusion:

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

(e)

To determine

The input energy and the change in internal energy for the block of ice as a system and $\Delta E_{\text{mech}}$ for the block-ice system.

Answer to Problem 20.76AP

The input energy is $0$ and the change in internal energy for the block of ice as a system is $5\text{J}$ and the change in mechanical energy for the block-ice system is $5\text{J}$.

Explanation of Solution

Given info: The mass of the copper block is $1.60\text{kg}$, the initial velocity of the copper block is $2.50\text{m}/\text{s}$, the final velocity of the block is $0\text{m}/\text{s}$, the temperature of the air is $0°\text{C}$, the temperature of the copper block is $0°\text{C}$ and the temperature of the ice is $0°\text{C}$.

The temperature of the block of the ice is $0°\text{C}$. The temperature is constant for block of the ice that means there is no temperature difference. There is no energy transfer by heat because there is no temperature difference for block of the ice. But the state of the block of the ice is changed when it melts, so the internal energy is not $0$.

From equation (1),  the expression for the change in internal energy.

$\Delta E_{\text{int}}=L\Delta m$

From part (a), the mass of the ice that melts is $15.0\text{mg}$.

Substitute $3.33\times {10}^5\text{J}/\text{kg}$ for $L$ and $15.0\text{mg}$ for $\Delta m$ in the above equation.

$\begin{array}{c}\Delta E_{\text{int}}=\left(3.33\times {10}^5\text{J}/\text{kg}\right)\left(15.0\text{mg}\times \frac{{10}^{-6}\text{kg}}{1\text{mg}}\right)\\ \approx 5\text{J}\end{array}$

Thus, the change in internal energy for the block of ice as a system is $5\text{J}$.

Write the expression for the change in mechanical energy for the block-ice system.

$\Delta E_{\text{mech}}=-\Delta K$

From part (a), change in kinetic energy of the block is $-5\text{J}$.

Substitute $-5\text{J}$ for $\Delta K$ in the above equation.

$\begin{array}{c}\Delta E_{\text{mech}}=-\left(-5\text{J}\right)\\ =5\text{J}\end{array}$

Conclusion:

Therefore, the input energy is $0$ and the change in internal energy for the block of ice as a system is $5\text{J}$ and the change in mechanical energy for the block-ice system is $5\text{J}$.

(f)

To determine

The input energy and the change in internal energy for the metal sheet as a system.

Answer to Problem 20.76AP

The input energy for the metal sheet as a system is $0$ and the change in internal energy for the metal sheet as a system is $0$.

Explanation of Solution

Given info: The mass of the copper block is $1.60\text{kg}$, the initial velocity of the copper block is $2.50\text{m}/\text{s}$, the final velocity of the block is $0\text{m}/\text{s}$, the temperature of the air is $0°\text{C}$, the temperature of the copper block is $0°\text{C}$ and the temperature of the sheet of the ice is $0°\text{C}$.

The temperature of the metal sheet is $0°\text{C}$. The temperature is constant for the metal sheet that means there is no temperature difference. There is no energy transfer by heat because there is no temperature difference for the metal sheet. The state of the metal sheet is constant that means there is no change of state of the metal sheet. So, the internal energy is also $0$ for the metal sheet because the state of the metal sheet is not changed.

Conclusion:

Therefore, the input energy for the metal sheet as a system is $0$ and the change in internal energy for the metal sheet as a system is $0$.

(g)

To determine

The change in temperature of both objects.

Answer to Problem 20.76AP

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

Explanation of Solution

Given info: The mass of the copper slab is $1.60\text{kg}$, the initial velocity of the copper slab is $2.50\text{m}/\text{s}$, the final velocity of the slab is $0\text{m}/\text{s}$ and the temperature of the copper slab is $20°\text{C}$.

Write the expression for the change in kinetic energy of the copper slab.

$\Delta K=\frac{1}{2}\left(\frac{1}{2}m\left(v_{\text{f}}^2-v_{\text{i}}^2\right)\right)$

Here,

$v_{\text{f}}$ is the final velocity of the copper slab.

$v_{\text{i}}$ is the initial velocity of the copper slab.

Substitute $0\text{m}/\text{s}$ for $v_{\text{f}}$, $2.50\text{m}/\text{s}$ for $v_{\text{i}}$ and $1.60\text{kg}$ for $m$ in the above equation.

$\begin{array}{c}\Delta K=\frac{1}{2}\left(\frac{1}{2}\left(1.60\text{kg}\right)\left({\left(0\text{m}/\text{s}\right)}^2-{\left(2.50\text{m}/\text{s}\right)}^2\right)\right)\\ =-2.5\text{J}\end{array}$

Thus, the change in kinetic energy of the copper slab is $-2.5\text{J}$.

Write the expression for the change in internal energy.

$\Delta K+\Delta E_{\text{int}}=0$

Substitute $-2.5\text{J}$ for $\Delta K$ in above equation.

$\begin{array}{c}-2.5\text{J}+\Delta E_{\text{int}}=0\\ \Delta E_{\text{int}}=2.5\text{J}\end{array}$

Thus, the change in internal energy of the copper slab is $2.5\text{J}$.

Write the expression for the change in internal energy due to the temperature change.

$\Delta E_{\text{int}}=mc\Delta T$

Here,

$m$ is the mass of the copper slab.

$c$ is the specific heat capacity.

$\Delta T$ is the change in temperature of both the objects.

Substitute $2.5\text{J}$ for $\Delta E_{\text{int}}$, $1.60\text{kg}$ for $m$ and $387\text{J}/\text{kg}\cdot °\text{C}$ for $c$ in above equation.

$\begin{array}{c}2.5\text{J}=\left(1.60\text{kg}\right)\left(387\text{J}/\text{kg}\cdot °\text{C}\right)\Delta T\\ \Delta T\approx 4.04\times {10}^{-3}°\text{C}\end{array}$

Conclusion:

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

(i)

To determine

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

Answer to Problem 20.76AP

The input energy is $0$ and the change in internal energy for the stationary slab is $2.5\text{J}$.

Explanation of Solution

Given info: The mass of the copper slab is $1.60\text{kg}$, the initial velocity of the copper slab is $2.50\text{m}/\text{s}$, the final velocity of the slab is $0\text{m}/\text{s}$ and the temperature of the copper slab is $20°\text{C}$.

The temperature of the stationary slab is $20°\text{C}$. The temperature is constant for the stationary slab that means there is no temperature difference. There is no energy transfer by heat because there is no temperature difference for the stationary slab. The kinetic energy of the stationary slab is converted into the internal energy due to the friction. From part (g), the change in internal energy of the sliding slab is $2.5\text{J}$. The internal energy of the stationary slab is also $2.5\text{J}$ because the temperature is same for both the slabs.

Conclusion:

Therefore, the input energy is $0$ and the change in internal energy for the stationary slab is $2.5\text{J}$.