## 08 February 2011

### Thermodynamics

Problem 1.
A rapidly spinning paddle wheel raises the temperature of 200mL of water from 21 degrees Celsius to 25 degrees. How much a) work is done and b) heat is transferred in this process?

Solution:
In this problem the work is done by the friction force. All the work will go to increase the internal energy of the water, which can be calculated as:
Where - specific heat of water, .

There are no heat transfer in the process.

Problem 2.
The temperature of a body is increased from -173 C to 357 C. What is the ratio of energies emitted by the body per second in these two cases?

Solution:
In this problem we need to use the Stefan–Boltzmann law: the amount of thermal radiation emitted per second by a (black) body is directly proportional to the fourth power of its absolute temperature:
where is a (Stefan-Boltzmann) constant. It is important that in the above expression the temperature is the absolute temperature (measured in Kelvin).
It means that
We convert the temperatures of the body into Kelvin:

Now we can find the ratio of the energies emitted by the body per second:

Problem 3.
How many btu are needed to change 10 pounds of ice at 5 degree Fahrenheit to steam at 250 degree Fahrenheit?

Solution:
The first step: we convert the temperatures given in Fahrenheit into the temperatures in Celsius:
5 degree Fahrenheit = -15 C
250 degree Fahrenheit = 121 C
Then we can see that initially we have m=10 pounds = 4.5 kg of ice and finally we have m = 4.5 kg of vapor. Therefore the whole process can be considered as the combination of the following processes:
(1) heating the ice from -15 C to 0 C;
(2) melting of ice at fixed temperature 0 C. After this process the ice becomes water.
(3) heating the water from 0 C to 100 C.
(4) vaporization of water at fixed temperature100 C. After this process the water becomes vapor.
(5) heating the vapor from 100 C to 121 C.

Process (1): the heat required to change the temperature of the ice is determined by the specific heat of ice:
The specific heat of ice is 2060 J/kg C. Then

Process (2): the heat required to melt the ice is determined by the specific latent heat of fusion:
The specific latent heat of fusion is 334000 J/kg. Then

Process (3): the heat required to change the temperature of the water is determined by the specific heat of water:
The specific heat of water is 4186 J/kg C. Then

Process (4): the heat required to vaporize the water is determined by the specific latent heat of vaporization:
The specific latent heat of vaporization is 2260000 J/kg. Then

Process (5): the heat required to change the temperature of the vapor is determined by the specific heat of vapor:
The specific heat of vapor is 1870 J/kg C. Then

Then the total heat is

We can convert the joule in btu:

Problem 4.
If ice has a density of , then what is the volume of 5,000 g of ice?

Solution:
The mass of ice is given in grams. The correct SI unit of the mass is kilogram (kg). We convert the mass of ice into kg:
Now we need to use the relation between the mass of the object ( ) , its density ( ), and its volume ( ):
From this expression we can find the volume of the ice:

Problem 5.
An ice cube having a mass of 50 grams and an initial temperature of -10 degrees Celsius is placed in 400 grams of 40 degrees Celsius water. What is the final temperature of the mixture if the effects of the container can be neglected?

Solution:
In this problem we need to use the energy conservation law. Namely, the energy transferred from the ice cube is equal to the energy transferred to the water.
Initially we have two systems: (1) ice cube and (2) water. The systems have different initial temperatures. In the final state the temperatures of the systems are the same – thermal equilibrium.
To find the final temperature we need to write the energy conservation law: the energy transferred from the system (1) [to the system (2)] is equal to the energy transferred to the system (2) [from the system (1)].
We introduce the final temperature of the systems: . We assume that the final temperature is greater than 0. It means that in the final state the system (1) (ice) becomes water. This is our assumption – if after the calculations we obtain that the temperature is less than 0, then we need to repeat the calculations with an assumption that the temperature is less than 0 or equals to 0.

Now we need to write the energy conservation law.
(I) energy (heat) transferred from system (1) has three contributions:
Process (1): the heat required to change the temperature of 50 g = 0.05 kg ice is determined by the specific heat of the ice:
The specific heat of ice is 2060 J/kg C, the initial temperature is (-10) and the final temperature is 0. Then the change of the temperature is 10:

Process (2): the heat required to melt the ice is determined by the specific latent heat of fusion:
The specific latent heat of fusion is 334000 J/kg. Then

Process (3): The ice now becomes the water and we need to increase the temperature of the water from 0 to the final temperature . The heat required to increase the temperature of the water is determined by the specific heat of the water (the mass of the water is equal to the mass of the ice):
The specific heat of the water is 4186 J/kg C. Then
Then the total heat transferred from the ice is

(II) energy (heat) transferred to the system (2) has only one contribution: We just need to decrease the temperature of the water from the initial temperature 40 degrees to the final temperature . The mass of the water is .
The heat required to decrease the temperature of the water is determined by the specific heat of the water:
The specific heat of the water is 4186 J/kg C, the change of the temperature of the water is . Then

Then the energy conservation takes the form:
From this equation we can find the final temperature of the system:

Problem 6.
Even if a man shows no visible perspiration he still evaporates about 500 grams of water per day from his lungs. How many calories of heat are removed by this evaporation? What is the rate of heat loss in watts due to this process?

Solution:
First we calculate the energy (heat) required to evaporate the water in SI units.
The SI unit of mass is kg. We need to convert 500 g in kg: .

The heat required to vaporize the water is determined by the specific latent heat of vaporization:
The specific latent heat of vaporization (in SI units) is 2260000 J/kg. Then

Now we can convert the energy (expressed in joules) into calories. We need to use the relation between joules and calories:
1 joule = 0.24 calories
Then

Now we can find the rate of heat loss (in watts) – the power. The rate of heat loss can be calculated as the heat loss per unit time. The watt is a SI unit. It means that we need to calculate the heat (energy) loss (in joule) per 1 second (SI unit of time).
We know that the energy is the energy loss per one day. To find the energy loss per 1 second we need to find the number of seconds in one day:

Problem 7.
One day the relative humidity is 90% and the temperature is 25 degrees Celsius. How many grams of water will condense out of each cubic meter of air if the temperature drops to 15 degrees Celsius? How many energy does the condensation from each cubic meter release?

Solution:
An air contains water vapor. The amount of water vapor, which a cubic meter of air can contain, cannot be more than a maximum value. We characterize the amount of water vapor in one cubic meter of air by its mass: . Then we can tell that the mass of the water vapor in one cubic meter of air should be less than the maximum value:
The maximum mass of the water vapor depends on the temperature: . If the vapor has the maximum mass then such vapor is called saturated vapor.
The humidity of the air is determined as the ratio of the mass of the vapor to the maximum possible vapor mass:

We know at 25 degrees Celsius the humidity is 90 %. It means that
From this expression we can find the mass of the vapor in the air:

The temperature of the air drops to 15 degrees Celsius. At this temperature the maximum possible mass of the vapor is . When the temperature of the air decreases the original mass of the vapor remains the same [ ]. If this mass is larger than the maximum mass then part of the vapor condenses into water. The mass of the water is
Therefore to find the mass of the water we need to find the maximum mass of the vapor.

We assume that the vapor is an ideal gas. If we introduce the maximum pressure of the vapor ( ) then the ideal gas equation takes the form (we write this equation for volume of 1 cubic meter)
Where - molar mass of water.
Then
The maximum pressures at 15 and 25 degrees Celsius are

Then the maximum masses of the vapor are

Now we can find the mass of the water

The second question: How many energy does the condensation from each cubic meter release?
The energy released from the condensation of the water is determined by the specific latent heat of vaporization:
The specific latent heat of vaporization (in SI units) is 2260000 J/kg. Then

# Ideal gas

Problem 1.
How much work has been done by an ideal gas, when the temperature of 5 moles of the gas increases by 2 Kelvin in an isobaric process?

Solution:
The process is isobaric. It means that the pressure is constant. Let us assume that the pressure is equal to and the volume of the ideal gas changes from initial value to the final value . Then by definition, the work done by the gas is
or
But we know that this is an ideal gas. It means that the equation of state of the gas is the Ideal Gas Law. This Ideal Gas Law is valid for both initial and the final states of the gas. Then
where is the number of moles of the gas. Then the work done by the gas is
We know that . Then
Problem 2.
What is an atomic mass of NaOH
Solution:
An atomic mass of a molecule NaOH is equal to the sum of
1. atomic mass of Na;
2. atomic mass of O;
3. atomic mass of H.
Atomic mass of Na is 22.99 u ( unified atomic mass unit ).
Atomics mass of O is 16 u.
Atomics mass of H is 1.008 u.
Then the atomic mass of NaOH is