The resistance of a platinum
wire at the ice and steam points
are 0.75 ohm and 1.05 ohm
respectively. Determine the
temperature at which the
resistance of the wire is 0.90
ohm?
A. 43.0 °C
B. 50.0°C
C. 69.9°C
D. 87.0°C
To determine the temperature at which the resistance of the wire is 0.90 ohm, we can use the formula for the resistance-temperature relationship of a material. This formula is:
R = R0 * (1 + α * (T - T0))
Where:
R is the resistance at temperature T
R0 is the resistance at a reference temperature T0
α is the temperature coefficient of resistance for the material
In this case, we have two data points:
R1 = 0.75 ohm at T1 (ice point)
R2 = 1.05 ohm at T2 (steam point)
We can use these points to determine the temperature coefficient of resistance (α) for platinum.
We can rearrange the formula to solve for α:
α = (R2 - R1) / (R1 * (T2 - T1))
Plugging in the values, we have:
α = (1.05 - 0.75) / (0.75 * (100 - 0)) = 0.4 / 75 = 0.0053 /°C
Now we can use this value of α to find the temperature (T) at which the resistance is 0.90 ohm:
0.90 = 0.75 * (1 + 0.0053 * (T - 0))
0.90 / 0.75 = 1 + 0.0053T
0.90 / 0.75 - 1 = 0.0053T
(0.90 - 0.75) / 0.75 = 0.0053T
0.15 / 0.75 = 0.0053T
0.20 = 0.0053T
T = 0.20 / 0.0053
T ≈ 37.74 °C
Therefore, the temperature at which the resistance of the wire is 0.90 ohm is approximately 37.74°C.
However, none of the given answer options match this value exactly.
To determine the temperature at which the resistance of the platinum wire is 0.90 ohm, we can use the concept of resistance-temperature relationship for metals.
The resistance-temperature relationship for metals can be expressed using the formula:
R = R0 * (1 + α * (T - T0))
Where:
R is the resistance at temperature T
R0 is the resistance at reference temperature T0
α is the temperature coefficient of resistance for the metal
In this case, we know the resistances at two different temperature points:
R1 = 0.75 ohm at T1 (the ice point)
R2 = 1.05 ohm at T2 (the steam point)
Using these values, we can calculate the temperature coefficient of resistance (α):
α = (R2 - R1) / (R1 * (T2 - T1))
Let's substitute the known values and solve for α:
α = (1.05 - 0.75) / (0.75 * (T2 - T1))
α = 0.30 / (0.75 * (T2 - T1))
α = 0.40 / (T2 - T1)
Now, we can use this value of α to find the temperature (T) at which the resistance (R) is 0.90 ohm.
0.90 = 0.75 * (1 + α * (T - T1))
0.90 = 0.75 * (1 + 0.40 / (T2 - T1) * (T - T1))
0.90 = 0.75 + 0.30 * (T - T1) / (T2 - T1)
0.15 = 0.30 * (T - T1) / (T2 - T1)
0.15 * (T2 - T1) = 0.30 * (T - T1)
0.15 * T2 - 0.15 * T1 = 0.30 * T - 0.30 * T1
0.15 * T2 + 0.30 * T1 = 0.30 * T + 0.15 * T1
0.15 * T2 + 0.15 * T1 = 0.30 * T
0.15 * (T2 + T1) = 0.30 * T
T2 + T1 = 2 * T
T = (T2 + T1) / 2
Now we can substitute the known values for T2 and T1 to find the temperature T:
T = (1.05 + 0.75) / 2
T = 0.9 / 2
T = 0.45
Therefore, the temperature at which the resistance of the platinum wire is 0.90 ohm is 0.45. However, this answer seems unlikely as it is below the ice point.
Therefore, the correct option may be missing or there might be an error in the information provided.