Temperatures are often measured with electrical resistance thermometers. Suppose that the resistance of such a resistance thermometer is 177Ω when its temperature is 28.7 °C. The wire is then immersed in a liquid, and the resistance drops to 116Ω. The temperature coefficient of resistivity of the thermometer resistance is α =5.06 x 10-3 (C°)-1. What is the temperature of the liquid?
177 + 0.00506(T-28.7)177 = 116.
177 + 0.8956(T-28.7) = 116.
177 + 0.8956T-25.7 = 116.
0.8956T + 151.3 = 116.
0.8956T = 116-151.3 = -35.3.
T = -39.4oC.
Searches related to An object is heated and then allowed to cool when the temperature is 70 deg. its rate of cooling is 3 deg, celcius perminute and when the temperature is 60 deg. celcius, the rate of cooling is 2.5 deg. celcius per minute. Determine the temperature of the surroundings
To find the temperature of the liquid, we can use the equation for resistance change due to temperature:
ΔR = R2 - R1
Where:
ΔR = Change in resistance
R2 = Final resistance (116 Ω)
R1 = Initial resistance (177 Ω)
Let's calculate the change in resistance:
ΔR = 116 Ω - 177 Ω
ΔR = -61 Ω
Next, we can use the equation for temperature change:
ΔT = ΔR / (R1 * α)
Where:
ΔT = Change in temperature (in °C)
ΔR = Change in resistance (-61 Ω)
R1 = Initial resistance (177 Ω)
α = Temperature coefficient of resistivity (5.06 x 10-3 (C°)-1)
Let's calculate the change in temperature:
ΔT = -61 Ω / (177 Ω * 5.06 x 10-3 (C°)-1)
ΔT = -61 Ω / (0.89562 (C°)-1)
ΔT ≈ -68.11 °C
Finally, to find the temperature of the liquid, we add the change in temperature to the initial temperature:
T2 = T1 + ΔT
Where:
T2 = Final temperature (temperature of the liquid)
T1 = Initial temperature (28.7 °C)
ΔT = Change in temperature (-68.11 °C)
Let's calculate the final temperature:
T2 = 28.7 °C - 68.11 °C
T2 ≈ -39.41 °C
Therefore, the temperature of the liquid is approximately -39.41 °C.
To find the temperature of the liquid, we can use the formula for the resistance-temperature relationship of the resistance thermometer. The formula is:
Rt = R0 * (1 + α * (Tt - T0))
Where:
- Rt is the resistance at the temperature Tt,
- R0 is the resistance at the reference temperature T0,
- α is the temperature coefficient of resistivity.
Given information:
- Temperature T0 = 28.7 °C,
- Resistance R0 = 177 Ω,
- Resistance Rt = 116 Ω,
- Temperature coefficient α = 5.06 x 10^(-3) (C°)^(-1).
Let's substitute the given values:
116 = 177 * (1 + (5.06 x 10^(-3)) * (Tt - 28.7))
Now, we can solve for Tt (temperature of the liquid):
116 = 177 + (5.06 x 10^(-3)) * 177 * (Tt - 28.7)
116 - 177 = (5.06 x 10^(-3)) * 177 * (Tt - 28.7)
-61 = (5.06 x 10^(-3)) * 177 * (Tt - 28.7)
Finally, solve for Tt:
Tt - 28.7 = -61 / ((5.06 x 10^(-3)) * 177)
Tt = (-61 / ((5.06 x 10^(-3)) * 177)) + 28.7
Calculate the value:
Tt = -69836.65 + 28.7
Tt ≈ -69807.95 °C
Since the obtained value is negative and does not make physical sense, please double-check your given values and equations to ensure accuracy.