What is the Continuous Flow of Energy Called
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3.6.2.1 Thermal energy transfer
Internal energy is the sum of the randomly distributed kinetic energies and potential energies of the particles in a body.
The internal energy of a system is increased when energy is transferred to it by heating or when work is done on it (and vice versa), eg a qualitative treatment of the first law of thermodynamics.
For a change of temperature:
$Q=mcΔθ$ where c is specific heat capacity.
Calculations including continuous flow.
This module, Thermal physics, focuses on the transfer of thermal energy and its effect on temperature and the state of the material. Although brief, the module contains quite a lot of maths, especially on this page, when discussing the transfer of heat, and also when deriving the equations of kinetic theory. Stick with it as the theory behind this module is pretty simple and the maths is surprisingly satisfying.
Heat and temperature
Heat and temperature are easily confused, but it is important to understand the difference. Two objects can have different amounts of heat, but have the same temperature, and vice-a-versa two object can have very different temperatures, but have the same amount of heat.
Heat can be thought of as flowing. If we put a hot object in thermal contact with a cold one, heat will flow from the hot object to the cold one. It never happens the other way around.
Heat is defined as the flow of energy from a warmer body to a cooler one. If there is no flow of energy there is very little sense of talking about heat. If we leave a hot object and a cold object out in a room which is at normal room temperature, there will be a flow of heat from the hot object into the room and a flow of heat from the room into the cold object.
So as heat flows from the cup of tea to the room and it loses heat its temperature drops and it will reach thermal equilibrium with the room when they are at the same temperature. The ice lolly will absorb heat from the room and will melt as it reaches thermal equilibrium. Two or more bodies are said to be in thermal equilibrium if they are at the same temperature and there is no net flow of energy between the them. Temperature difference can be thought of as the driving force behind heat flow.
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Internal Energy
Inside any solid, liquid or gas at any temperature above absolute zero, the particles are constantly moving; in a gas the particles move freely in random directions, in a solid they vibrate about a fixed position. Therefore the particles have kinetic energy and potential energy due to their inter-molecular bonds. The more the particles vibrate or move, the hotter the material. The internal energy of the material is the sum of the distribution of the kinetic energies of the particles in the material and the potential energies. Temperature is the measure of the internal energy of a substance (Internal energy is sometimes called thermal energy as magnetised metals also have higher internal energies).The particles move with a range of kinetic energies so in a material such as a gas, where there are no inter-molecular bonds the temperature is directly proportional to the average kinetic energy. The internal energy of a substance can be changed by,
- Transfer of heat or energy.
- Doing work on the object.
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Specific Heat Capacity
If two buckets of water with equal masses and different temperatures are mixed together they will reach a thermal equilibrium. The cooler mass of water will increase in temperature and the warmer mass will decrease. As they are the same mass of the same material, they will share out their heat equally. In other words they will do an equal amount of work on each other and there will be an equal amount of heat flow between them. So as in the picture below, they will come to a thermal equilibrium at a temperature which is half of the temperature difference between them. In this case the equilibrium temperature is $\quantity{50}{° C}{}{}$.
If one of the buckets of water has a larger mass of water and a higher temperature it will be able to do more work on the cooler bucket and therefore more heat will be able to flow to the cooler bucket and it will be able to increase its temperature by a larger amount. Again, because they are both the same substance the amount of heat flow is proportional to only the temperature difference and the mass. The heat (Q1 ) lost by the hotter water is equal to the heat (Q2 ) gained by the cooler water.
$$Q_{1} = Q_{2} $$
The temperature difference between the two buckets of water is $\quantity{60}{° C}{}{}$ and the larger bucket contains $\frac{3}{4}$ of the total mass of water. The cooler bucket will absorb the heat from the warmer water and they will come to thermal equilibrium. As the hotter bucket has less mass it also has less heat, so it cannot increase the temperature of the more massive cooler bucket by much so the equilibrium temperature is closer to the initial temperature of the cooler bucket at $\quantity{35}{° C}{}{}$
$$temperature\;change\;=\;temperature\;difference\;\times\;(1 - proportion\;of\;total\;mass)$$ $$\Delta\theta=\theta_{1}-\theta_{2} \times \left(1-\frac{1}{4} \right )=\quantity {15}{^{\circ}C}{}{}$$
However different materials can absorb different amounts of heat for the same temperature rise, even when their masses are equal. The amount of heat that needs to be supplied to increase them temperature of $\quantity{1}{kg}{}{}$ of substance by $\quantity{1}{° C}{}{}$ is called its specific heat capacity and is defined by the equation:
$$\large Q=mc\Delta\theta$$
Where:
- Q is the heat supplied in joules (J)
- m is the mass in kilograms (kg)
- c is the specific heat capacity in J kg-1 °C-1
- Δθ is the change in temperature in °C or kelvin (K)
Water, for example has a very high specific heat capacity of $\quantity {4200}{J}{kg^{-1}}{° C^{-1}}$ which means that and $\quantity{1}{kg}{}{}$ bucket of water would require $\quantity{4200}{J}{}{}$ of energy to increase its temperature by a single degree whereas copper which has a lower SHC of only $\quantity {385}{J}{kg^{-1}}{° C^{-1}}$ which means that the same mass of material experiencing the same amount of heat transfer would increase its temperature by nearly 11 degree.
We can still apply a similar logic as above when mixing two materials with different SHCs and different temperatures. The temperature difference will drive the flow of heat, and the two materials will come to a thermal equilibrium at the same temperature, even though they contain different amounts of heat. However, the material with the higher SHC will change temperature less than an equal mass of material with a lower SHC.
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Worked example
An example of this is an simple experiment to find the temperature of a Bunsen burner flame. A $\quantity{50}{g}{}{}$ brass mass (mb ) is heated on a gauze above the flame for several minutes, until it reaches thermal equilibrium with the flame (θI ). It is then quickly dropped into a bucket containing $\quantity{250}{g}{}{}$ of water (mw ) at $\quantity{23}{° C}{}{}$ (θi ). The brass mass needs to be dropped quickly to stop it radiating some if its heat away and to prevent too much of the water evaporating as it is introduced to the bucket. The water quickly rises in temperature to $\quantity{33}{° C}{}{}$ (θF ) as the two material reach thermal equilibrium. The initial temperature of the brass, and therefore the flame can be estimated by realising that the heat supplied by the brass (Q1 ) is equal to the heat received by the water (Q2 )
$$Q_{1}=Q_{2}\\ m_{b}c_{b}\left(\theta_{I}-\theta_{F}\right)=m_{w}c_{w}\left(\theta_{F}-\theta_{i} \right )$$
The specific heat capacity of water (cw ) is $\quantity{4200}{J}{kg^{-1}}{° C^{-1}}$ and the specific heat capacity of brass (cb ) is $\quantity{380}{J}{kg^{-1}}{° C^{-1}}$. As the only unknown is the initial temperature of the brass (θI ) the equation can be rearranged and solved:
\begin{align} m_{b}c_{b}\left(\theta_{I}-\theta_{F}\right) & =m_{w}c_{w}\left(\theta_{F}-\theta_{i} \right )\\ \\ \theta_{I} & =\left( \frac{m_{w}c_{w}\left(\theta_{F}-\theta_{i} \right )}{m_{b}c_{b}}\right) +\theta_{F}\\ \\ \theta_{I}&= \left( \frac{\quantity{0.5}{kg}{}{} \times \quantity{4200}{J}{kg^{-1}}{° C^{-1}}\times \left(\quantity{33}{° C}{}{}-\quantity{23}{° C}{}{} \right )}{\quantity{0.05}{kg}{}{} \times \quantity{380}{J}{kg^{-1}}{° C^{-1}}}\right) +\quantity{33}{° C}{}{}\\ \\ \theta_{I}&=\quantity{586}{{°} C}{}{}\;\left(\mathrm{2\; sig.\; fig.}\right) \end{align}
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Finding the specific heat capacity of materials experimentally.
In class you will carry out two experiments to determine the specific heat capacity of water and different metals. The first, simple method, shown below is similar a known mass of water, or metal is heated by either a submerged resistor on an immersion heater. If the current (I) and the potential difference (V) across the heater are known, then the power can be calculated. The initial temperature (θi ) of the material is recorded.
The apparatus are left so that the water or metal is heated for 10 minutes. The p.d. and current are recorded every minute so that the energy delivered by the heater can be calculated more accurately. The final temperature (θf ) is also recorded and the specific heat capacity can be calculated simply. This method gives fairly accurate results, but suffers from two main problems, firstly if the material under test is not well insulated, heat is lost readily to the environment, so more heat is required to be supplied to the material giving higher than expected results. Also as the insulation or container will also absorb some of the heat, so its specific heat capacity should also considered. Fortunately the second method manages to account for these heat losses.
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Measuring the SHC of a material using the continuous flow method
Water is made to flow through a tube containing an electric heating element. The tube is surrounded by an evacuated container to reduce an unnecessary heat losses through conduction and convection. The heater is connected to a power supply which is a adjusted to give a temperature rise (Δθ=θ2-θ1 ) of a few degrees across the length of the tube. Once a steady state has been achieved the mass (m) of water passing through the tube across a period of time (t) is measured. The flow rate through the tube is then changed and the heater is adjusted so that Δθ is the same as previously. This allows the SHC to be calculated and the heat losses to the environment through radiation to be eliminated.
electrical energy supplied in time t = energy transferred to fluid + energy lost to surroundings
So:
$$\large IVt=mcΔθ+E$$
Where
- I is the current through the heater in A
- V is the p.d. across the heater in V
- t is the time in s for the mass m of fluid to flow
For the second flow rate m2 in time t2 , I and V are adjusted to allow Δθ to remain the same as before. Now:
$$I_{2}V_{2}t_{2}=m_{2}cΔθ+E$$
As the temperature difference Δθ is the same as before, the amount of heat radiated to the surroundings is also the same. Therefore E can be eliminated by subtracting these two equations from each other.
$$IVt-I_{2}V_{2}t_{2}=mcΔθ-m_{2}cΔθ$$
As c is the only unknown in the equation it can be calculated by factorising the right-hand side to give:
$$IVt-I_{2}V_{2}t_{2}=cΔθ\left(m-m_{2}\right)$$
$$c=\frac{IVt-I_{2}V_{2}t_{2}}{Δθ\left(m-m_{2}\right)}$$
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Resources
Source: http://www.alevelphysicsnotes.com/thermal_physics/specific_heat.php
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