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 : Thermodynamics

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

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

Possible Answers:

\(\displaystyle 10.7C\)

\(\displaystyle 214C\)

\(\displaystyle 984C\)

\(\displaystyle 49.2C\)

\(\displaystyle 119.2C\)

Correct answer:

\(\displaystyle 119.2C\)

Explanation:

This is a simple pairing of two \(\displaystyle Q = mc\Delta T\) equations.

\(\displaystyle mc\Delta T = mc\Delta T\)

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

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

Solving for \(\displaystyle \Delta T\) of water yields \(\displaystyle 49.2C\), and certainly the water gained heat not lost it, so the final temperature of the water is \(\displaystyle 70C+49.2C = 119.2C\)

Example Question #1 : Thermodynamics

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

Possible Answers:

\(\displaystyle 1.092 mm\)

\(\displaystyle 0.92 mm\)

\(\displaystyle 1.32 mm\)

\(\displaystyle 2.02 mm\)

Correct answer:

\(\displaystyle 1.092 mm\)

Explanation:

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

\(\displaystyle \Delta L = \alpha L_o \Delta T\)

We are given:

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

\(\displaystyle L_o = 7.0 m\) for the initial length of the steel.

We need to calculate \(\displaystyle \Delta T\) which is the the difference of the two temperatures.

\(\displaystyle \Delta T = 38-25 = 13\)

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

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

Example Question #1 : Thermodynamics

What is the change in entropy for \(\displaystyle 1kg\) of ice initially at \(\displaystyle 0C\) that melts to water and warms to the temperature of the atmosphere at \(\displaystyle 25C\)?

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

Specific heat of water: \(\displaystyle 4190 \frac{J}{kg\cdot K}\)

Specific heat of ice: \(\displaystyle 2100 \frac{J}{kg\cdot K}\) 

Possible Answers:

\(\displaystyle 1220 \frac{J}{ K}\)

\(\displaystyle 1590 \frac{J}{K}\)

\(\displaystyle 890 \frac{J}{K}\)

\(\displaystyle 370 \frac{J}{K}\)

Correct answer:

\(\displaystyle 1590 \frac{J}{K}\)

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:

\(\displaystyle \Delta S = \frac{\Delta Q}{T}\)

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

\(\displaystyle \Delta S = \frac{mL_f}{T}\)

Now we plug in our known values into the equation:

Mass \(\displaystyle m=1kg\)

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

Temperature \(\displaystyle T=273.15K\)

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

  • Now we need to calculate the change in entropy for the liquid water warming from \(\displaystyle 0C\) to \(\displaystyle 25C\).

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

\(\displaystyle \Delta S = \int_{T_o}^{T_f} \frac{\Delta Q}{T}\)

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

\(\displaystyle \Delta S = \int_{T_o}^{T_f}\frac{mc\Delta T}{T}\)

next we integrate this equation to get:

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

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

Final temperature of water \(\displaystyle T_f = 298.15 K\)

Initial temperature of water \(\displaystyle T_o = 273.15 K\)

Specific heat of water \(\displaystyle c = 4190 \frac{J}{K}\)

Mass of water \(\displaystyle m = 1 kg\)

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

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

\(\displaystyle 1222.7\frac{J}{K} + 366.9\frac{J}{K} \approx 1590\frac{J}{K}\)

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