<span>In the </span>natural logarithm<span> format or in equivalent notation (see: </span>logarithm) as:
base<span> e</span><span> assumed, is called the </span>Planck entropy<span>, </span>Boltzmann entropy<span>, Boltzmann entropy formula, or </span>Boltzmann-Planck entropy formula<span>, a </span>statistical mechanics<span>, </span><span> </span>S<span> is the </span>entropy<span> of an </span>ideal gas system<span>, </span>k<span> is the </span>Boltzmann constant<span> (ideal </span>gas constant R<span> divided by </span>Avogadro's number N<span>), and </span>W<span>, from the German Wahrscheinlichkeit (var-SHINE-leash-kite), meaning probability, often referred to as </span>multiplicity<span> (in English), is the number of “</span>states<span>” (often modeled as quantum states), or "complexions", the </span>particles<span> or </span>entities<span> of the system can be found in according to the various </span>energies<span> with which they may each be assigned; wherein the particles of the system are assumed to have uncorrelated velocities and thus abide by the </span>Boltzmann chaos assumption<span>.
I hope this helps. </span>
It would go to the positive charge im like 99% sure of that
Answer:
The final equilibrium T_{f} = 25.7[°C]
Explanation:
In order to solve this problem we must have a clear concept of heat transfer. Heat transfer is defined as the transmission of heat from one body that is at a higher temperature to another at a lower temperature.
That is to say for this case the heat is transferred from the iron to the water, the temperature of the water will increase, while the temperature of the iron will decrease. At the end of the process a thermal balance is found, i.e. the temperature of iron and water will be equal.
The temperature of thermal equilibrium will be T_f.
The heat absorbed by water will be equal to the heat rejected by Iron.
Heat transfer can be found by means of the following equation.
where:
Qiron = Iron heat transfer [kJ]
m = iron mass = 200 [g] = 0.2 [kg]
T_i = Initial temperature of the iron = 300 [°C]
T_f = final temperature [°C]
Cp_iron = 437 [J/kg*°C]
Cp_water = 4200 [J/kg*°C]
![0.2*437*(300-T_{f})=1*4200*(T_{f}-20)\\26220-87.4*T_{f}=4200*T_{f}-84000\\26220+84000=4200*T_{f}+87.4*T_{f}\\110220 = 4287.4*T_{f}\\T_{f}=25.7[C]](https://tex.z-dn.net/?f=0.2%2A437%2A%28300-T_%7Bf%7D%29%3D1%2A4200%2A%28T_%7Bf%7D-20%29%5C%5C26220-87.4%2AT_%7Bf%7D%3D4200%2AT_%7Bf%7D-84000%5C%5C26220%2B84000%3D4200%2AT_%7Bf%7D%2B87.4%2AT_%7Bf%7D%5C%5C110220%20%3D%204287.4%2AT_%7Bf%7D%5C%5CT_%7Bf%7D%3D25.7%5BC%5D)
A - proximal convoluted tubule
3.75kmper squ hr
acc = (v - u) /t
=(54-32)/8
=22/8
= 3.75kmhr^-2
