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Chapter 1: Problem 5

The resistance of a platinum wire is found to be \(11.000 \Omega\) at the ice point, \(15.247 \Omega\) at the steam point, and \(28.887 \Omega\) at the sulfur point. Find the constants \(a\) and \(b\) in the cquation $$ R^{\prime}=R_{0}^{\prime}\left(1+a \theta+b \theta^{2}\right) $$ and plot \(R^{\prime}\) against Celsius temperature \(\theta\) in the range from 0 to \(660^{\circ} \mathrm{C}\).

Short Answer

Expert verified
a = 0.00385, b = -0.000007; use \(R' = 11.000 (1 + 0.00385 \theta - 0.000007 \theta^2)\) to plot resistance over 0 to 660°C.

Step by step solution

01

- Understand the Variables

Let the resistance at the ice point (0°C) be denoted as \(R_0 = 11.000 \Omega\), at the steam point (100°C) as \(R_{100} = 15.247 \Omega\), and at the sulfur point (444.6°C) as \(R_{444.6} = 28.887 \Omega\).
02

- Set Up the Equation for Known Temperatures

The general equation for resistance is: \( R = R_0 (1 + a \theta + b \theta^2) \). Using the known points: At 0°C: \( R_0 = 11.000 \Omega \) At 100°C: \( R_{100} = 15.247 \Omega \) At 444.6°C: \( R_{444.6} = 28.887 \Omega \).
03

- Write Equations for 100°C and 444.6°C

For 100°C: \( 15.247 = 11.000 (1 + 100a + 10000b) \).For 444.6°C: \( 28.887 = 11.000 (1 + 444.6a + 197672.16b) \).
04

- Solve for Constants a and b

From the first equation: \(15.247 = 11.000 + 1100a + 110000b \)This simplifies to: \( 4.247 = 1100a + 110000b \).From the second equation: \(28.887 = 11.000 + 48806a + 2174393.76b \)This simplifies to: \(17.887 = 48806a + 2174393.76b \).Now solve these two linear equations to find \(a\) and \(b\).
05

- Perform Calculation

By solving the linear equations, use substitution or matrix method:From: \( 4.247 = 1100a + 110000b \)From: \( 17.887 = 48806a + 2174393.76b \)Simplifying and solving for \(a\) and \(b\) gives approximately:a = 0.00385b = -0.000007.
06

- Plot the Resistance against Temperature

Using the values obtained for \(a\) and \(b\), plot the curve using the equation:\( R' = 11.000 (1 + 0.00385 \theta - 0.000007 \theta^2) \).Evaluate this equation for various values of \( \theta \) ranging from 0 to \(660^\text{°}\text{C} \) and then plot the resulting points.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistance vs Temperature
Understanding the relationship between resistance and temperature is crucial in the study of materials like platinum. Platinum resistance thermometers are widely used because platinum has a predictable change in resistance with temperature. At 0°C, the resistance of platinum is denoted as \(R_0\) and can be measured. With increasing temperature, the resistance \(R\) increases, and this relationship is often expressed mathematically to allow accurate temperature measurements.
Linear and Quadratic Coefficients
In the equation \[ R = R_0 (1 + a \theta + b \theta^2) \], the coefficients \(a\) and \(b\) play key roles:
  • \(a\) is the linear coefficient, representing how resistance changes linearly with temperature.
  • \(b\) is the quadratic coefficient, representing how resistance changes with the square of the temperature, addressing non-linear behavior.
To find these coefficients, we use known resistances at specific temperatures. For example, in the given exercise, resistances at 0°C, 100°C, and 444.6°C are used to form equations. Solving these equations simultaneously determines the values of \(a\) and \(b\).
Plotting and Graphing Data
Plotting the relationship between resistance and temperature helps visualize how resistance increases with temperature:
  • Step 1: Calculate resistance values for a range of temperatures using the determined equation.
  • Step 2: Plot these resistance values against corresponding temperatures on a graph.
  • Step 3: Connect the points to see the curve formed by the equation.
In our case, we use \[ R' = 11.000 (1 + 0.00385 \theta - 0.000007 \theta^2) \] to calculate and plot resistance for temperatures from 0 to 660°C. This plotted curve provides a clear visual of the platinum wire’s behavior over the extensive temperature range.
Solving Linear Equations
To find the constants \(a\) and \(b\), solving linear equations is essential. Here’s the process simplified:
  • Write down the given resistances and temperatures as equations involving \(a\) and \(b\).
  • From the exercise, we have two equations: \[ 4.247 = 1100a + 110000b \] and \[ 17.887 = 48806a + 2174393.76b \]
  • Use substitution or matrix methods to solve these equations.
As shown, solving these provides values for \(a\) and \(b\), essential for accurately describing the resistance-temperature relationship. In this instance, \(a \approx 0.00385\) and \(b \approx -0.000007\).

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Most popular questions from this chapter

In the table below, a number in the top row represents the pressure of a gas in the bulb of a constant-volume gas thermometer (corrected for dead space, thermal expansion of bulb, etc.) when the bulb is immersed in a water triple- point cell. The bottom row represents the corresponding readings of pressure when the bulb is surrounded by a material at a constant unknown temperature. Calculate the idealgas temperature \(T\) of this material. (Use five significant figures.) \(P_{T P}, \mathrm{kPa} \quad 133.32\) \(\begin{array}{lll}99.992 & 66.661 & 33.331\end{array}\) \(P, \mathrm{kPa}\) \(204.69 \quad 153.54\) \(102.37\) \(51.190\)

The resistance \(R^{\prime}\) of a particular carbon resistor obeys the equation $$ \sqrt{\frac{\log R^{\prime}}{T}}=a+b \log R^{\prime} $$ where \(a=-1.16\) and \(b=0.675\). (a) In a liquid helium cryostat, the resistance is found to be exactly \(1000 \Omega\) (ohms). What is the temperature? (b) Make a log-log graph of \(R^{\prime}\) against \(T\) in the resistance range from 1000 to \(30,000 \Omega\)

What is the temperature on the Fahrenheit scale of the normal boiling point of \(\mathrm{H}_{2} \mathrm{O}\). if this temperature is \(99.974^{\circ} \mathrm{C} ?\) (Use five significant figures.)

The limiting value of the ratio of pressures of a gas at the steam point and at the triple point of water when the gas is kept at constant volume is found to be \(1.365954\). What is the ideal-gas temperature of the steam point to six significant figures?
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