Math Problem Statement

  1. The mass of an object ie found to be (24.5 +_ 0.2)g. A. Find: Absulute uncertainty (Au) B Find : Relative uncertainty (Ru II. The density of an object was found to be 2.7 g/cubic cm +_ 1.5%. What is the absolute uncertainty of the density of the object? III. The following are the values obtained in measuring the temperature of a room: 31.0 degree celcius, 30.5 C, 30.0 C, 31.3 C, 30.8 C, 30.6 C, and 31.1 Celcius. What is the room temperature? Show the step by step procedure in getting the uncertainty. A. Find: Mean= B. Find: Range= C. Find: Uncertainty = D. What is the room temperature?

Solution

Let's go through each part of your question step by step.


I. MASS OF THE OBJECT: (24.5 ± 0.2) g

A. Find Absolute Uncertainty (Au):

  • The absolute uncertainty is the given uncertainty in the measurement, which is already provided as ± 0.2 g.

Thus,
Au = 0.2 g

B. Find Relative Uncertainty (Ru):

  • The relative uncertainty is the ratio of the absolute uncertainty to the measured value, typically expressed as a percentage.

Ru=Absolute Uncertainty (Au)Measured Value×100\text{Ru} = \frac{\text{Absolute Uncertainty (Au)}}{\text{Measured Value}} \times 100

Ru=0.224.5×1000.82%\text{Ru} = \frac{0.2}{24.5} \times 100 \approx 0.82\%

Thus,
Ru ≈ 0.82%


II. DENSITY OF THE OBJECT: 2.7 g/cm³ ± 1.5%

You are given the relative uncertainty as 1.5%, and we need to calculate the absolute uncertainty.

Absolute Uncertainty (Au) in density can be calculated as:

Au=(Relative Uncertainty100)×Measured Value\text{Au} = \left( \frac{\text{Relative Uncertainty}}{100} \right) \times \text{Measured Value}

Au=(1.5100)×2.7=0.0405g/cm3\text{Au} = \left( \frac{1.5}{100} \right) \times 2.7 = 0.0405 \, \text{g/cm}^3

Thus,
Absolute Uncertainty (Au) = 0.0405 g/cm³


III. ROOM TEMPERATURE MEASUREMENTS

Given temperature values:
31.0°C, 30.5°C, 30.0°C, 31.3°C, 30.8°C, 30.6°C, and 31.1°C

A. Find Mean Temperature: To find the mean, add up all the measurements and divide by the number of measurements.

Mean=31.0+30.5+30.0+31.3+30.8+30.6+31.17\text{Mean} = \frac{31.0 + 30.5 + 30.0 + 31.3 + 30.8 + 30.6 + 31.1}{7}

\text{Mean} = \frac{215.3}{7} \approx 30.76 \,^\circ C

Thus,
Mean ≈ 30.76°C

B. Find Range: The range is the difference between the maximum and minimum values.

Maximum value = 31.3°C
Minimum value = 30.0°C

\text{Range} = 31.3 - 30.0 = 1.3 \,^\circ C

Thus,
Range = 1.3°C

C. Find Uncertainty: The uncertainty can be estimated as half of the range.

\text{Uncertainty} = \frac{\text{Range}}{2} = \frac{1.3}{2} = 0.65 \,^\circ C

Thus,
Uncertainty = 0.65°C

D. What is the Room Temperature? The room temperature is the mean value with the uncertainty.

Thus,
Room Temperature = (30.76 ± 0.65)°C


Summary of Results:

  1. Absolute Uncertainty (Au) of mass = 0.2 g
  2. Relative Uncertainty (Ru) of mass ≈ 0.82%
  3. Absolute Uncertainty (Au) of density ≈ 0.0405 g/cm³
  4. Mean Room Temperature30.76°C
  5. Range = 1.3°C
  6. Uncertainty = 0.65°C
  7. Room Temperature(30.76 ± 0.65)°C

Let me know if you'd like further clarification on any part or have any questions!

Related Questions:

  1. How do you calculate relative uncertainty if given an absolute error?
  2. Why is range used to estimate uncertainty in repeated measurements?
  3. What is the significance of uncertainty in scientific measurements?
  4. Can you convert relative uncertainty to absolute uncertainty in any context?
  5. How do you interpret the final room temperature with its uncertainty?

Tip:

In measurements, always express the final result with its uncertainty to reflect the precision of your data.

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Math Problem Analysis

Mathematical Concepts

Uncertainty in Measurement
Relative Uncertainty
Absolute Uncertainty
Mean
Range
Basic Statistics

Formulas

Absolute Uncertainty (Au) = ± value in the measurement
Relative Uncertainty (Ru) = (Absolute Uncertainty / Measured Value) × 100
Mean = Sum of values / Number of values
Range = Max value - Min value
Uncertainty = Range / 2

Theorems

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Suitable Grade Level

Grades 10-12