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College Physics : Laws of Thermodynamics and General Concepts

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Example Question #1 :College Physics

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:

10.7C

119.2C

49.2C

214C

984C

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 #2 :College Physics

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:

1.092 mm

1.32 mm

2.02 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 #3 :College Physics

What is the change in entropy for 1kg of ice initially atthat 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:

$890 \frac{J}{K}$

$1590 \frac{J}{K}$

$370 \frac{J}{K}$

$1220 \frac{J}{ K}$

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 aconstant temperaturewhen 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, andis 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 fromto.

The basic formula to calculate the change in entropy at achanging temperaturewhen 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:

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

接下来,我们代入我们的值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

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