Thermodynamics and Heat Transfer

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Thermodynamics and Heat Transfer

In thermodynamics, the basic equation for the change in internal energy in a system is:

$\Delta E=Q+W~$

$\Delta E~$ is the change in the internal energy on the system in Joules ($J$).

$Q~$ is the heat flow INTO the system in Joules ($J$).

$W~$ is the work done ON the system in Joules ($J$).

It's important to be aware of the sign conventions for $Q$ $W$. A positive $Q$ corresponds to heat flow into the system from its surroundings, while a negative $Q$ corresponds to heat flowing out of the system to its surroundings. A positive $W$ corresponds to work being done on the system by the surroundings, while a negative $W$ corresponds to work being done by the system on its surroundings. These conventions make sense since if heat flows INTO a system, the internal energy should go up, if heat flows OUT of a system (or dissipates into the environment) the internal energy should go down. If the work is done BY the system, then the system must have used up its own energy to do the work, and so the internal energy will decrease. If work is done ON the system, then something outside is putting energy into it, so the internal energy increases.

For heat transfer from a hot to cold thermal resevoir, we have the equation:

${\frac {dQ}{dt}}={\frac {kA(T_{h}-T_{c})}{L}}$

$dQ/dt$ is the rate of heat flow (it can also be written $Q/t$) in Watts i.e. Joules per second ($W$ i.e. $J/s$).

$k$ is the thermal conductivity in Watts per meter-Kelvin ($W/mK$).

$A$ is the area that is in contact in metres squared ($m^{2}$).

$T_{h}$ is the temperature of the hot thermal reservoir in Kelvin ($K$)

$T_{c}$ is the temperature of the cold thermal resevoir in Kelvin ($L$)

$L$ is the thickness of the object transferring the heat in meters ($m$).

Note that although the standard units for temperature are Kelvin, the common units for temperature, degrees Celcius, only differ from Kevlin by a constant term. So if we have a difference between two temperatures, we can use either Kevlin or Celcius, since the constant term (273) will be cancelled out when we subtract ((25 + 273) - (10 + 273) = (25 - 10) = 15).

This equation describes how fast heat energy flows between two objects which can be considered resevoirs (their temperature doesn't change perceptively as heat flows in or out). If we have, for exapmle, a rod conducting heat, then the $A$ refers to the cross-sectional area of the rod and $L$ refers to the thickness. If we have some large object, like a house, then the $A$ refers to all of the area in contact between the two heat reservoirs, in this case the area of the walls and the ceiling (and perhaps the floor).

Ideal Gas

An ideal gas is a model for gasses with very small molecular volume and negligible inter-particle and intra-particle interations. The equation of state for and ideal gas is called the ideal gas law and is written:

$PV=NkT~$

$P$ is the pressure in Pascals ($Pa$).

$V$ is the volums in metres cubed ($m^{3}$).

$N$ is the number of particles.

$k$ is Boltzmann's Constant, approximately $1.38\times 10^{-23}$ Joules per Kelvin ($J/K$).

$T$ is temperature in Kelvin ($K$).

The ideal gas law is sometimes also written in terms of the ideal gas constant:

$PV=nRT~$

In this case $n$ is the number of moles (One mole is $6.02\times 10^{23}$ particles).

$R$ is the ideal gas constant, or $(6.02\times 10^{23})$ times Boltzman's Constant, which is approximately $8.31$ Joules per mole-Kelvin ($J/molK$).

Temperature is a measure of the average kinetic energy of particles in a system. Since, for an ideal gas, it is assumed that all internal energy is kinetic energy, there is a very simple formula relating the energy of an ideal gas to its temperature:

$E={\frac {3}{2}}NkT={\frac {3}{2}}nRT$

Problem Solving

Q A sealed cylinder contains a sample of ideal gas at pressure of 2atm. The rms speed of the molecules is v. If this rmv speed is then reduced to 0.70v, what will be the new pressure of the gas?

A Remember that the internal energy of an ideal gas is all kinetic energy, so:

$E={\frac {3}{2}}nRT={\frac {1}{2}}mv^{2}$

$\rightarrow nRT={\frac {1}{3}}mv^{2}$

Substitute this into the ideal gas law:

$P_{1}V=nRT={\frac {1}{3}}mv^{2}$

$\rightarrow P_{1}={\frac {1}{3}}{\frac {mv^{2}}{V}}$

Then we can replace $v$ with $(0.70v)$ to find $P_{2}$

$P_{2}={\frac {1}{3}}{\frac {m(0.70v)^{2}}{V}}=0.49{\frac {1}{3}}{\frac {mv^{2}}{V}}=0.49P_{1}=0.49(2atm)=0.98atm$

Q In a system, the gas in the cylinder is at a pressure of 1.01*10^5 Pa and the piston has an area of 0.100 m2. As energy is slowly added to the gas by heat, the piston is pushed up a distance of 4.00 cm. Calculate the work done by the expanding gas on the surroundings, Wenv, assuming the pressure remains constant.

A The work done on the environment is the negative of the work done on the gas, so compute the change in volume and multiply by the pressure:

$(deltaV)=(A)(deltaY)=(0.100m^{2})(4.00*10^{2}m)=4.00*10^{3}m^{3}$

$Wenv=(P)(deltaV)=(1.01*10^{5}Pa)(4.00*10^{3}m^{3})=404J$

Q How much thermal energy must be added to 5.00 moles of monatomic ideal gas at 3.00 * 10^2 K and with a constant volume of 1.50 L in order to raise the temperature of the gas by 3.80 * 10^2 K?

A The energy transferred by heat is equal to the change in the internal energy of the gas, which can be calculated by substitution into Equation Q = (n)(C)(deltaT):

$Q=deltaU=(n)(C)(deltaT)=(3/2)nR(deltaT)=(3/2)(5.00mol)(8.31J/K.mol)(80.0k)=4.99*10^{3}J$

Black Body Radiation

A blackbody emits radiations at every wavelength (usually described in form of a black body radiation curve). However, the peak radiation wavelength decreases with temperature, and is described by Wien's Law:

$\lambda _{peak}={\frac {b}{T}}$

$\lambda _{peak}~$ is the peak wavelength in metres ($m$).

$b\approx 2.898\times 10^{-3}mK$

$T~$ is the temperature in Kelvin ($K$).

As temperature increases, the total energy emitted by the black body increases. The power radiated is described by Stefan's Law:

$P=\sigma AT^{4}~$

$P~$ is the power radiated in Watts ($W$).

$\sigma \approx 5.67\times 10^{-8}{\frac {W}{m^{2}K^{4}}}$

$A~$ is surface area of the body in metres squared ($m^{2}$).

$T~$ is the temperature of the body in Kelvin ($K$).

Problem Solving

Q: Given that a $300K$ blackbody radiates its peak energy at a wavelength of about $\lambda _{cold}=10.0\mu m$, at what wavelength $\lambda _{hot}$ would a $600K$ blackbody radiate its peak energy?

A: We know from Wien's Law that $\lambda _{peak}=b/T$, where $\lambda _{peak}$ is the peak wavelength of the radiation, and $T$ is the surface temperature. Rearrange the equation to get:

$\lambda _{peak}T=b~$

When there are two sets of temperatures, we have:

$\lambda _{1}T_{1}=b~$

and

$\lambda _{2}T_{2}=b~$

thus

$\lambda _{1}T_{1}=\lambda _{2}T_{2}~$

So the wavelength of the $600K$ blackbody is

$(300K)(10\mu m)=(600K)\lambda _{2}~$

$\rightarrow \lambda _{2}=5\mu m$

Other formula that might be useful include Planck Energy Distribution Formula which calculates the power output at each temperature P_l = [2*pi*h*c^c]/l^5 (e^K - I) where K = (hc/lkT). P_l is the power intensity, h is Planck's constant = 6.626*10^{-34}, c = speed of light, I = wavelength, k = Boltzmann constant and T = temperature.

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