Question 392: Current Electricity

Question

Resistance of a wire at $$0^{\circ} \mathrm{C}, 100^{\circ} \mathrm{C}$$ and $$t^{\circ} \mathrm{C}$$ is found to be $$10 \Omega, 10.2 \Omega$$ and $$10.95 \Omega$$ respectively. The temperature $$t$$ in Kelvin scale is _________.

Accepted Answer

To determine the temperature $$t$$ in the Kelvin scale, we need to use the relationship between the resistance of a wire and temperature. The general formula for the resistance $R$ of a wire as a function of temperature is:
$$ R_t = R_0 (1 + \alpha t) $$
where:
$$R_t$$ is the resistance at temperature $$t$$
$$R_0$$ is the resistance at the reference temperature (usually $$0^{\circ} \mathrm{C}$$)
$$\alpha$$ is the temperature coefficient of resistance
$$t$$ is the temperature
We are given the following resistances:
Resistance at $$0^{\circ} \mathrm{C}$$: $$R_0 = 10 \Omega$$
Resistance at $$100^{\circ} \mathrm{C}$$: $$R_{100} = 10.2 \Omega$$
Resistance at $$t^{\circ} \mathrm{C}$$: $$R_t = 10.95 \Omega$$
First, we need to find the temperature coefficient of resistance $$\alpha$$. Using the resistance at $$100^{\circ} \mathrm{C}$$:
$$ 10.2 = 10 (1 + \alpha \cdot 100) $$
Solving for $$\alpha$$:
$$ \frac{10.2}{10} = 1 + 100\alpha \implies 1.02 = 1 + 100\alpha \implies 100\alpha = 0.02 \implies \alpha = \frac{0.02}{100} = 0.0002 $$
Now, we can find the temperature $$t$$ using the resistance at $$t^{\circ} \mathrm{C}$$:
$$ 10.95 = 10 (1 + 0.0002 \cdot t) $$
Solving for $$t$$:
$$ \frac{10.95}{10} = 1 + 0.0002 t \implies 1.095 = 1 + 0.0002 t \implies 0.0002 t = 0.095 \implies t = \frac{0.095}{0.0002} = 475 $$
The temperature $$t$$ in Celsius is $$475^{\circ} \mathrm{C}$$. To convert this to the Kelvin scale:
$$ T_{K} = t_{C} + 273.15 = 475 + 273.15 = 748.15 \, \mathrm{K} $$
So, the temperature $$t$$ in Kelvin scale is approximately $$748.15 \, \mathrm{K}$$.

Current Electricity

Resistance of a wire at $0^{\circ} \mathrm{C}, 100^{\circ} \mathrm{C} and t^{\circ} \mathrm{C} is found to be 10 \Omega, 10.2 \Omega and 10.95 \Omega respectively. The temperature t$ in Kelvin scale is _________.

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