A reistance thermometer has a resistance of 20w at 0°c and 85w at 100°c if it resistance is 52w in medium calculate the corresponding temperature?
if the response is linear, then
0 + (52-20)/(85-20) * (100-0) = 640/13 = 49.23°C
makes sense.
52 is just about halfway from 20 to 85,
and 49.23 is just about halfway from 0 to 100.
y = m x + b :)
w = (65/100) T + 20
52 = 0.65 T + 20
0.65 T = 32
etc
Who can tutor physics from Africa
any physicist in Africa, be my guess ...
answer
To find the corresponding temperature for a given resistance in the medium, we can use linear interpolation.
Linear interpolation is a method to estimate a value between two known values based on a linear relationship. In this case, we have two known resistance-temperature pairs: (20Ω, 0°C) and (85Ω, 100°C).
The formula for linear interpolation is:
temperature = temperature1 + ((resistance - resistance1) / (resistance2 - resistance1)) * (temperature2 - temperature1),
where:
temperature1 = 0°C,
temperature2 = 100°C,
resistance1 = 20Ω,
resistance2 = 85Ω,
resistance = 52Ω.
Now, let's plug the values into the formula and calculate the corresponding temperature:
temperature = 0°C + ((52Ω - 20Ω) / (85Ω - 20Ω)) * (100°C - 0°C)
= 0°C + (32Ω / 65Ω) * 100°C
= 0°C + 0.492308 * 100°C
≈ 49.23°C.
Therefore, the corresponding temperature for a resistance of 52Ω in the medium is approximately 49.23°C.