Laws of Thermodynamics and General Concepts - Physics

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Question

A steel sword is cooled from to by dipping it in of water that is at .

If the specific heat of the steel is $502$\frac{J}{kg\cdot C}$$ and the specific heat of water is $4187$\frac{J}{kg\cdot C}$$ , what will be the temperature of the water after the steel is cooled?

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Answer

This is a simple pairing of two $Q = mc\Delta T$ equations.

$mc\Delta T = mc\Delta T$

We will make the left side for the steel and the right side for water. Thus:

$(20)(502)(500-90) = (20)(4187)\Delta T$

Solving for $\Delta T$ of water yields , and certainly the water gained heat not lost it, so the final temperature of the water is

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Question

A steel beam is long at a temperature of . On a hot day, the temperature reaches . What is the change in the beam's length due to thermal expansion given that the thermal expansion coefficient for steel is $1.2\times $10^{-5}$ ?$

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Answer

We need to use the equation for thermal expansion in order to solve this problem:

$\Delta L = \alpha L_o \Delta T$

We are given:

$\alpha = 1.2\times $10^{-5}$$ for the thermal expansion constant

for the initial length of the steel.

We need to calculate $\Delta T$ which is the the difference of the two temperatures.

$\Delta T = 38-25 = 13$

Now we have enough information to solve for the change in the length of steel.

$\Delta L = $(1.2\times10^{-5}$)(7)(13) = 0.001092m = 1.092mm$

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Question

What is the change in entropy for of ice initially at that melts to water and warms to the temperature of the atmosphere at ?

Latent heat of fusion for water:

Specific heat of water: $4190 $\frac{J}{kg\cdot K}$$

Specific heat of ice: $2100 $\frac{J}{kg\cdot K}$$

Tap to reveal answer

Answer

We must account for two separate processes since the ice is melting and then warming. The basic formula to calculate the change in entropy at a constant temperature when the ice melts is:

$\Delta S = $\frac{\Delta Q}{T}$$

Where $\Delta S$ is the change in entropy, $\Delta Q$ is the change in heat energy, and is the temperature at which the change of state for the ice occurs. Now we substitute in an equation for $\Delta Q$ for when ice melts: $\Delta Q = mL_f$ . Our entropy equation now becomes:

$\Delta S = $\frac{mL_f}{T}$$

Now we plug in our known values into the equation:

Mass

Latent heat of fusion $L_f=3.34\times $10^{-5}$$\frac{J}{kg}$$

Temperature

And we get the change in entropy for the ice melting at a constant temperature, which is $1222.7$\frac{J}{K}$$

Now we need to calculate the change in entropy for the liquid water warming from to .

The basic formula to calculate the change in entropy at a changing temperature when the water warms is:

$\Delta S = $\int_{T_$o}$^{T_f}$ $\frac{\Delta Q}{T}$$

Next we substitute in an equation for $\Delta Q$ when there is a temperature change, but not a change in state. That equation is: $\Delta Q = mc\Delta T$ so our entropy equation now becomes:

$\Delta S = $\int_{T_$o}$^{T_f}$$\frac{mc\Delta T}{T}$$

next we integrate this equation to get:

$\Delta S = mc \ln($\frac{T_f}{T_o}$)$

Next we plug in the values we know and solve for the change in entropy:

Final temperature of water

Initial temperature of water

Specific heat of water $c = 4190 $\frac{J}{K}$$

Mass of water

Now we get the change in entropy for the water warming to be $366.9 $\frac{J}{K}$$

We add our two entropy values together to find the total change in entropy to be:

$1222.7$\frac{J}{K}$ + 366.9$\frac{J}{K}$ \approx 1590$\frac{J}{K}$$

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Question

A steel sword is cooled from to by dipping it in of water that is at .

If the specific heat of the steel is $502$\frac{J}{kg\cdot C}$$ and the specific heat of water is $4187$\frac{J}{kg\cdot C}$$ , what will be the temperature of the water after the steel is cooled?

Tap to reveal answer

Answer

This is a simple pairing of two $Q = mc\Delta T$ equations.

$mc\Delta T = mc\Delta T$

We will make the left side for the steel and the right side for water. Thus:

$(20)(502)(500-90) = (20)(4187)\Delta T$

Solving for $\Delta T$ of water yields , and certainly the water gained heat not lost it, so the final temperature of the water is

← Didn't Know|Knew It →

Question

A steel beam is long at a temperature of . On a hot day, the temperature reaches . What is the change in the beam's length due to thermal expansion given that the thermal expansion coefficient for steel is $1.2\times $10^{-5}$ ?$

Tap to reveal answer

Answer

We need to use the equation for thermal expansion in order to solve this problem:

$\Delta L = \alpha L_o \Delta T$

We are given:

$\alpha = 1.2\times $10^{-5}$$ for the thermal expansion constant

for the initial length of the steel.

We need to calculate $\Delta T$ which is the the difference of the two temperatures.

$\Delta T = 38-25 = 13$

Now we have enough information to solve for the change in the length of steel.

$\Delta L = $(1.2\times10^{-5}$)(7)(13) = 0.001092m = 1.092mm$

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Question

What is the change in entropy for of ice initially at that melts to water and warms to the temperature of the atmosphere at ?

Latent heat of fusion for water:

Specific heat of water: $4190 $\frac{J}{kg\cdot K}$$

Specific heat of ice: $2100 $\frac{J}{kg\cdot K}$$

Tap to reveal answer

Answer

We must account for two separate processes since the ice is melting and then warming. The basic formula to calculate the change in entropy at a constant temperature when the ice melts is:

$\Delta S = $\frac{\Delta Q}{T}$$

Where $\Delta S$ is the change in entropy, $\Delta Q$ is the change in heat energy, and is the temperature at which the change of state for the ice occurs. Now we substitute in an equation for $\Delta Q$ for when ice melts: $\Delta Q = mL_f$ . Our entropy equation now becomes:

$\Delta S = $\frac{mL_f}{T}$$

Now we plug in our known values into the equation:

Mass

Latent heat of fusion $L_f=3.34\times $10^{-5}$$\frac{J}{kg}$$

Temperature

And we get the change in entropy for the ice melting at a constant temperature, which is $1222.7$\frac{J}{K}$$

Now we need to calculate the change in entropy for the liquid water warming from to .

The basic formula to calculate the change in entropy at a changing temperature when the water warms is:

$\Delta S = $\int_{T_$o}$^{T_f}$ $\frac{\Delta Q}{T}$$

Next we substitute in an equation for $\Delta Q$ when there is a temperature change, but not a change in state. That equation is: $\Delta Q = mc\Delta T$ so our entropy equation now becomes:

$\Delta S = $\int_{T_$o}$^{T_f}$$\frac{mc\Delta T}{T}$$

next we integrate this equation to get:

$\Delta S = mc \ln($\frac{T_f}{T_o}$)$

Next we plug in the values we know and solve for the change in entropy:

Final temperature of water

Initial temperature of water

Specific heat of water $c = 4190 $\frac{J}{K}$$

Mass of water

Now we get the change in entropy for the water warming to be $366.9 $\frac{J}{K}$$

We add our two entropy values together to find the total change in entropy to be:

$1222.7$\frac{J}{K}$ + 366.9$\frac{J}{K}$ \approx 1590$\frac{J}{K}$$

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Question

A steel sword is cooled from to by dipping it in of water that is at .

If the specific heat of the steel is $502$\frac{J}{kg\cdot C}$$ and the specific heat of water is $4187$\frac{J}{kg\cdot C}$$ , what will be the temperature of the water after the steel is cooled?

Tap to reveal answer

Answer

This is a simple pairing of two $Q = mc\Delta T$ equations.

$mc\Delta T = mc\Delta T$

We will make the left side for the steel and the right side for water. Thus:

$(20)(502)(500-90) = (20)(4187)\Delta T$

Solving for $\Delta T$ of water yields , and certainly the water gained heat not lost it, so the final temperature of the water is

← Didn't Know|Knew It →

Question

A steel beam is long at a temperature of . On a hot day, the temperature reaches . What is the change in the beam's length due to thermal expansion given that the thermal expansion coefficient for steel is $1.2\times $10^{-5}$ ?$

Tap to reveal answer

Answer

We need to use the equation for thermal expansion in order to solve this problem:

$\Delta L = \alpha L_o \Delta T$

We are given:

$\alpha = 1.2\times $10^{-5}$$ for the thermal expansion constant

for the initial length of the steel.

We need to calculate $\Delta T$ which is the the difference of the two temperatures.

$\Delta T = 38-25 = 13$

Now we have enough information to solve for the change in the length of steel.

$\Delta L = $(1.2\times10^{-5}$)(7)(13) = 0.001092m = 1.092mm$

← Didn't Know|Knew It →

Question

What is the change in entropy for of ice initially at that melts to water and warms to the temperature of the atmosphere at ?

Latent heat of fusion for water:

Specific heat of water: $4190 $\frac{J}{kg\cdot K}$$

Specific heat of ice: $2100 $\frac{J}{kg\cdot K}$$

Tap to reveal answer

Answer

We must account for two separate processes since the ice is melting and then warming. The basic formula to calculate the change in entropy at a constant temperature when the ice melts is:

$\Delta S = $\frac{\Delta Q}{T}$$

Where $\Delta S$ is the change in entropy, $\Delta Q$ is the change in heat energy, and is the temperature at which the change of state for the ice occurs. Now we substitute in an equation for $\Delta Q$ for when ice melts: $\Delta Q = mL_f$ . Our entropy equation now becomes:

$\Delta S = $\frac{mL_f}{T}$$

Now we plug in our known values into the equation:

Mass

Latent heat of fusion $L_f=3.34\times $10^{-5}$$\frac{J}{kg}$$

Temperature

And we get the change in entropy for the ice melting at a constant temperature, which is $1222.7$\frac{J}{K}$$

Now we need to calculate the change in entropy for the liquid water warming from to .

The basic formula to calculate the change in entropy at a changing temperature when the water warms is:

$\Delta S = $\int_{T_$o}$^{T_f}$ $\frac{\Delta Q}{T}$$

Next we substitute in an equation for $\Delta Q$ when there is a temperature change, but not a change in state. That equation is: $\Delta Q = mc\Delta T$ so our entropy equation now becomes:

$\Delta S = $\int_{T_$o}$^{T_f}$$\frac{mc\Delta T}{T}$$

next we integrate this equation to get:

$\Delta S = mc \ln($\frac{T_f}{T_o}$)$

Next we plug in the values we know and solve for the change in entropy:

Final temperature of water

Initial temperature of water

Specific heat of water $c = 4190 $\frac{J}{K}$$

Mass of water

Now we get the change in entropy for the water warming to be $366.9 $\frac{J}{K}$$

We add our two entropy values together to find the total change in entropy to be:

$1222.7$\frac{J}{K}$ + 366.9$\frac{J}{K}$ \approx 1590$\frac{J}{K}$$

← Didn't Know|Knew It →

Question

A steel sword is cooled from to by dipping it in of water that is at .

If the specific heat of the steel is $502$\frac{J}{kg\cdot C}$$ and the specific heat of water is $4187$\frac{J}{kg\cdot C}$$ , what will be the temperature of the water after the steel is cooled?

Tap to reveal answer

Answer

This is a simple pairing of two $Q = mc\Delta T$ equations.

$mc\Delta T = mc\Delta T$

We will make the left side for the steel and the right side for water. Thus:

$(20)(502)(500-90) = (20)(4187)\Delta T$

Solving for $\Delta T$ of water yields , and certainly the water gained heat not lost it, so the final temperature of the water is

← Didn't Know|Knew It →

Question

A steel beam is long at a temperature of . On a hot day, the temperature reaches . What is the change in the beam's length due to thermal expansion given that the thermal expansion coefficient for steel is $1.2\times $10^{-5}$ ?$

Tap to reveal answer

Answer

We need to use the equation for thermal expansion in order to solve this problem:

$\Delta L = \alpha L_o \Delta T$

We are given:

$\alpha = 1.2\times $10^{-5}$$ for the thermal expansion constant

for the initial length of the steel.

We need to calculate $\Delta T$ which is the the difference of the two temperatures.

$\Delta T = 38-25 = 13$

Now we have enough information to solve for the change in the length of steel.

$\Delta L = $(1.2\times10^{-5}$)(7)(13) = 0.001092m = 1.092mm$

← Didn't Know|Knew It →

Question

What is the change in entropy for of ice initially at that melts to water and warms to the temperature of the atmosphere at ?

Latent heat of fusion for water:

Specific heat of water: $4190 $\frac{J}{kg\cdot K}$$

Specific heat of ice: $2100 $\frac{J}{kg\cdot K}$$

Tap to reveal answer

Answer

We must account for two separate processes since the ice is melting and then warming. The basic formula to calculate the change in entropy at a constant temperature when the ice melts is:

$\Delta S = $\frac{\Delta Q}{T}$$

Where $\Delta S$ is the change in entropy, $\Delta Q$ is the change in heat energy, and is the temperature at which the change of state for the ice occurs. Now we substitute in an equation for $\Delta Q$ for when ice melts: $\Delta Q = mL_f$ . Our entropy equation now becomes:

$\Delta S = $\frac{mL_f}{T}$$

Now we plug in our known values into the equation:

Mass

Latent heat of fusion $L_f=3.34\times $10^{-5}$$\frac{J}{kg}$$

Temperature

And we get the change in entropy for the ice melting at a constant temperature, which is $1222.7$\frac{J}{K}$$

Now we need to calculate the change in entropy for the liquid water warming from to .

The basic formula to calculate the change in entropy at a changing temperature when the water warms is:

$\Delta S = $\int_{T_$o}$^{T_f}$ $\frac{\Delta Q}{T}$$

Next we substitute in an equation for $\Delta Q$ when there is a temperature change, but not a change in state. That equation is: $\Delta Q = mc\Delta T$ so our entropy equation now becomes:

$\Delta S = $\int_{T_$o}$^{T_f}$$\frac{mc\Delta T}{T}$$

next we integrate this equation to get:

$\Delta S = mc \ln($\frac{T_f}{T_o}$)$

Next we plug in the values we know and solve for the change in entropy:

Final temperature of water

Initial temperature of water

Specific heat of water $c = 4190 $\frac{J}{K}$$

Mass of water

Now we get the change in entropy for the water warming to be $366.9 $\frac{J}{K}$$

We add our two entropy values together to find the total change in entropy to be:

$1222.7$\frac{J}{K}$ + 366.9$\frac{J}{K}$ \approx 1590$\frac{J}{K}$$

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