Question

exercise 6: measuring water temperature in a beaker 22.1 °c, 22.0 °c, 22.2 °c, 40.0 °c a) find the average: b) find the range: c) find the random error: d) write the result with the error: e) percentage of uncertainty:

Explanation:

Part a)

Step1: Sum the temperatures

Sum = \( 22.1 + 22.0 + 22.2 + 40.0 = 106.3 \)

Step2: Divide by number of values (4)

Average = \( \frac{106.3}{4} = 26.575 \, ^\circ\text{C} \)

Step1: Identify max and min

Max = \( 40.0 \, ^\circ\text{C} \), Min = \( 22.0 \, ^\circ\text{C} \)

Step2: Subtract min from max

Range = \( 40.0 - 22.0 = 18.0 \, ^\circ\text{C} \)

Step1: For random error (half of range, for symmetric data)

Random Error = \( \frac{\text{Range}}{2} = \frac{18.0}{2} = 9.0 \, ^\circ\text{C} \) (Note: Alternatively, if excluding the outlier 40.0, range of 22.0,22.1,22.2 is 0.2, error 0.1. But as per given data, we use all 4 values here.)

Answer:

\( 26.575 \, ^\circ\text{C} \)

Part b)