Platinium Resistance Thermometer

In summary, the conversation discussed using a platinum resistance thermometer to interpolate temperatures between the freezing point of water and 700.0 degrees Celsius. The temperature can be calculated using the formula R = R,o(1 + A Tc + B Tc^2), where R,o, A, and B are constants determined by measurements at the ice point, steam point, and melting point of zinc. Using experimental data, the constants were calculated to be R,o = 7.85630 ohms, A = 0.0736, and B = -1.209x10^(-5), which showed a slight departure from linearity over the relevant temperature range.
  • #1
peachtree40
1
0

Homework Statement


In the interval between the freezing point (ice point) of water and 700.0deg-C, a platinum resistance thermometer is to be used for interpolating temperatures from 0 °C to the melting point of zinc, 692.666 K. The temperature, in Celcius or Centigrade, is given by a formula for the variation of the resistance of the thermometer with temperature: R = R,o(1 + A Tc + B Tc^2). Ro, A, and B are constants determined by measurements at the ice point, the steam point, and the melting point of zinc. If R equals 7.85630 ohms at the ice point, 65.56082 ohms at the steam point, and 233.60848 ohms at zinc's melting point, find Ro.[/B]

Homework Equations

The Attempt at a Solution


(0, 7.85630) (100, 65.56080) (692.666-273.15, 233.608)
These are three ordered pairs of data for the equation
I found Ro by just plugging in 0 and finding that Ro is 7.85630
I then plugged in the other ordered pairs and used algebra to eliminate A
65.56082=7.8563+7.8563(100A)+7.8563(10000B)
57.7045=785.63(A+10B)
.07345=A+10B (used for elimination)

233.608=7.8563+7.8563(419.514A)+7.8563(175992B)
225.7417=3295.83(A+419.514B)
.0685= A +419.514B (used for elimination)

after subtracting the two equations to find B and cancel A I find that B= -1.209x10^(-5)
then plugging back into one of the shorter equations above find A to be .0736, but both of these are supposedly wrong...
 
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  • #2
Approach appears valid.

There is no 'supposedly' about it - you obtained your constants from the experimental points. When you feed these constants and the temperatures back into the equation they must give you back your original experimental points. Anything else must be due to an algebraic or arithmetic mistake.
That is the test, not anything anyone says.

I get constants not wildly different from yours so I think this is basically OK.
They give a visible but not great departure from linearity over the relevant range.

More to add on the science of this if the OP comes back.
 
Last edited:

Related to Platinium Resistance Thermometer

What is a Platinium Resistance Thermometer?

A Platinium Resistance Thermometer is a type of thermometer that uses the electrical resistance of a thin wire made of platinum to measure temperature. It is highly accurate and commonly used in scientific and industrial applications.

How does a Platinium Resistance Thermometer work?

A Platinium Resistance Thermometer works by passing an electrical current through a thin wire made of platinum. As the temperature changes, the resistance of the wire also changes, allowing the thermometer to accurately measure the temperature.

What are the advantages of using a Platinium Resistance Thermometer?

There are several advantages to using a Platinium Resistance Thermometer, including high accuracy, stability, and repeatability. They also have a wide temperature range and are resistant to corrosion and oxidation.

How accurate is a Platinium Resistance Thermometer?

Platinium Resistance Thermometers are highly accurate, with a typical accuracy of ±0.03°C at 0°C. They also have a high resolution, making them suitable for precision measurements.

What are the common applications of Platinium Resistance Thermometers?

Platinium Resistance Thermometers are commonly used in scientific research, industrial processes, and environmental monitoring. They are also used in medical and pharmaceutical settings, as well as in food and beverage production.

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