A platinum-resistance thermometer has a resistance of 5Ω at 0°C and 9Ω at l00°C. Assuming that resistance changes uniformly with temperature, calculate the ...
Question 1 Report
A platinum-resistance thermometer has a resistance of 5Ω at 0°C and 9Ω at l00°C. Assuming that resistance changes uniformly with temperature, calculate the resistance of the thermometer when the temperature is 45°C.
Answer Details
A platinum-resistance thermometer has a resistance of 5 Ω at 0°C and 9 Ω at 100°C, and it is assumed that resistance changes uniformly with temperature. We need to calculate the resistance of the thermometer when the temperature is 45°C.
We can start by finding the change in resistance as the temperature changes from 0°C to 100°C:
ΔR = 9 Ω - 5 Ω = 4 Ω
This change occurs over a temperature range of 100°C - 0°C = 100°C, so the resistance changes by 4 Ω per 100°C.
To find the resistance at 45°C, we can use the following proportion:
(ΔR / ΔT) = (R₂ - R₁) / (T₂ - T₁)
Where:
ΔR = change in resistance
ΔT = change in temperature
R₁ = initial resistance (at 0°C)
R₂ = final resistance (at 45°C)
T₁ = initial temperature (0°C)
T₂ = final temperature (45°C)
Substituting the known values, we get:
(4 Ω / 100°C) = (R₂ - 5 Ω) / (45°C - 0°C)
Simplifying the equation:
R₂ - 5 Ω = (4 Ω / 100°C) x 45°C
R₂ - 5 Ω = 1.8 Ω
R₂ = 6.8 Ω
Therefore, the resistance of the platinum-resistance thermometer at 45°C is 6.8 Ω. Answer: 6.8 Ω.
Download The App On Google Playstore
Everything you need to excel in your exams
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning