1. The resistance of the element of a platinum resistance thermometer is 2.00 Ω at the ice point and 2.73 Ω at the steam point. What temperature on the platinum resistance scale wo... 1. The resistance of the element of a platinum resistance thermometer is 2.00 Ω at the ice point and 2.73 Ω at the steam point. What temperature on the platinum resistance scale would correspond to a resistance of 8.43 Ω? 2. If the electrical resistance from a platinum at 0°C is 10.5 Ω and 12.8 Ω at 100°C. Calculate the temperature when the resistance is 11.75 Ω.
Understand the Problem
The problem presents two scenarios involving platinum resistance thermometers. In the first scenario, you're given the resistance at the ice point and steam point and are asked to find the temperature corresponding to a given resistance. In the second scenario, you are given the electrical resistance from a platinum at 0°C and 100°C, and are asked to find the temperature corresponding to a different given resistance. This involves understanding the relationship between resistance and temperature for platinum resistance thermometers and applying linear interpolation.
Answer
(1) $880.82 \, ^\circ C$ (2) $54.35 \, ^\circ C$
Answer for screen readers
(1) The temperature is $880.82 , ^\circ C$. (2) The temperature is $54.35 , ^\circ C$.
Steps to Solve
- Problem 1: Find the change in Resistance per degree Celsius
The resistance changes linearly with temperature. The ice point is 0°C and the steam point is 100°C. So, calculate the change in resistance per degree Celsius:
$ \Delta R = 2.73 , \Omega - 2.00 , \Omega = 0.73 , \Omega $
$ \Delta T = 100^\circ C - 0^\circ C = 100^\circ C $
$ Change , in , Resistance , per , ^\circ C = \frac{\Delta R}{\Delta T} = \frac{0.73 , \Omega}{100^\circ C} = 0.0073 , \Omega/^\circ C $
- Problem 1: Find the temperature corresponding to 8.43 $\Omega$
Calculate the change in resistance from the ice point ($2.00 , \Omega$) to the given resistance ($8.43 , \Omega$):
$ Change , in , Resistance = 8.43 , \Omega - 2.00 , \Omega = 6.43 , \Omega $
Now, divide this change in resistance by the change in resistance per degree Celsius to find the temperature:
$ Temperature = \frac{Change , in , Resistance}{Change , in , Resistance , per , ^\circ C} = \frac{6.43 , \Omega}{0.0073 , \Omega/^\circ C} = 880.82 , ^\circ C $
- Problem 2: Find the Change in Resistance per degree Celsius
The resistance changes linearly with temperature. We are given: Resistance at 0°C ($R_0$) = $10.5 , \Omega$ Resistance at 100°C ($R_{100}$) = $12.8 , \Omega$
Calculate the change in resistance: $ \Delta R = R_{100} - R_0 = 12.8 , \Omega - 10.5 , \Omega = 2.3 , \Omega $ Calculate the change in temperature: $ \Delta T = 100^\circ C - 0^\circ C = 100^\circ C $ Calculate change in resistance per degree Celsius: $ Change , in , Resistance , per , ^\circ C = \frac{\Delta R}{\Delta T} = \frac{2.3 , \Omega}{100^\circ C} = 0.023 , \Omega/^\circ C $
- Problem 2: Find the temperature corresponding to 11.75 $\Omega$
Calculate the change in resistance from 0°C ($10.5 , \Omega$) to the given resistance ($11.75 , \Omega$):
$ Change , in , Resistance = 11.75 , \Omega - 10.5 , \Omega = 1.25 , \Omega $
Now, divide this change in resistance by the change in resistance per degree Celsius to find the temperature:
$ Temperature = \frac{Change , in , Resistance}{Change , in , Resistance , per , ^\circ C} = \frac{1.25 , \Omega}{0.023 , \Omega/^\circ C} = 54.35 , ^\circ C $
(1) The temperature is $880.82 , ^\circ C$. (2) The temperature is $54.35 , ^\circ C$.
More Information
Platinum resistance thermometers rely on the principle that the electrical resistance of platinum changes predictably with temperature. This allows for accurate temperature measurement over a wide range. The relationship between resistance and temperature is nearly linear, which simplifies calculations.
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