Example Question #1 : Thermodynamics
A steel sword is cooled from to by dipping it in of water that is at .
If the specific heat of the steel is and the specific heat of water is , what will be the temperature of the water after the steel is cooled?
This is a simple pairing of two equations.
We will make the left side for the steel and the right side for water. Thus:
Solving for of water yields , and certainly the water gained heat not lost it, so the final temperature of the water is
Example Question #2 : Thermodynamics
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
We need to use the equation for thermal expansion in order to solve this problem:
We are given:
for the thermal expansion constant
for the initial length of the steel.
We need to calculate which is the the difference of the two temperatures.
Now we have enough information to solve for the change in the length of steel.
Example Question #3 : Thermodynamics
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:
Specific heat of ice:
- 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:
Where is the change in entropy, 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 for when ice melts: . Our entropy equation now becomes:
Now we plug in our known values into the equation:
Latent heat of fusion
And we get the change in entropy for the ice melting at a constant temperature, which is
- 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:
Next we substitute in an equation for when there is a temperature change, but not a change in state. That equation is: so our entropy equation now becomes:
next we integrate this equation to get:
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
Mass of water
Now we get the change in entropy for the water warming to be
We add our two entropy values together to find the total change in entropy to be: