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College Physics : Laws of Thermodynamics and General Concepts

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Example Question #1 : Laws Of Thermodynamics And General Concepts

A 20kg steel sword is cooled from 500C to 90C by dipping it in 20kg of water that is at 70C .

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?

Possible Answers:

49.2C

984C

119.2C

214C

10.7C

Correct answer:

119.2C

Explanation:

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 49.2C , and certainly the water gained heat not lost it, so the final temperature of the water is 70C+49.2C = 119.2C

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Example Question #1 : Thermodynamics

A steel beam is 7.0 m long at a temperature of 25C . On a hot day, the temperature reaches 38C . What is the change in the beam's length due to thermal expansion given that the thermal expansion coefficient for steel is $1.2\times 10^{-5} ?$

Possible Answers:

2.02 mm

1.092 mm

1.32 mm

0.92 mm

Correct answer:

1.092 mm

Explanation:

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

L_o = 7.0 m 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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Example Question #1 : Thermodynamics

What is the change in entropy for 1kg of ice initially at that melts to water and warms to the temperature of the atmosphere at 25C ?

Latent heat of fusion for water: $L_{fw} = 3.34 \times 10^{-5} \frac{J}{kg}$

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

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

Possible Answers:

$370 \frac{J}{K}$

$1220 \frac{J}{ K}$

$890 \frac{J}{K}$

$1590 \frac{J}{K}$

Correct answer:

$1590 \frac{J}{K}$

Explanation:

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 m=1kg

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

Temperature T=273.15K

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 T_f = 298.15 K

Initial temperature of water T_o = 273.15 K

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

Mass of water m = 1 kg

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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College Physics : Laws of Thermodynamics and General Concepts

Study concepts, example questions & explanations for College Physics

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Example Questions

Example Question #1 : Laws Of Thermodynamics And General Concepts

Example Question #1 : Thermodynamics

Example Question #1 : Thermodynamics

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