NCERT Solutions

We need to determine the final steady-state temperature of a nichrome heating element using its resistance-temperature relationship.

Concept: Temperature coefficient of resistance

The electrical resistance of a conductor varies with temperature according to the temperature coefficient of resistance ($\alpha$). The relationship is given by:

$$R = R_0 [1 + \alpha(T - T_0)]$$

where $R$ is the resistance at temperature $T$, and $R_0$ is the resistance at reference temperature $T_0$. We also use Ohm's Law, $V = IR$, to find the resistance from the given voltage and current.

Given:

• Supply voltage, $V = 230\,\text{V}$
• Initial current at room temperature, $I_1 = 3.2\,\text{A}$
• Room temperature, $T_1 = 27.0\,^{\circ}\text{C}$
• Steady-state current, $I_2 = 2.8\,\text{A}$
• Temperature coefficient of nichrome, $\alpha = 1.70 \times 10^{-4}\,^{\circ}\text{C}^{-1}$

Solving:

1. Calculate the initial resistance ($R_1$) at room temperature ($T_1$):

$$\begin{aligned} R_1 &= \frac{V}{I_1} \\ &= \frac{230\,\text{V}}{3.2\,\text{A}} \\ &= 71.875\,\Omega \end{aligned}$$

2. Calculate the steady-state resistance ($R_2$) at temperature $T_2$:

$$\begin{aligned} R_2 &= \frac{V}{I_2} \\ &= \frac{230\,\text{V}}{2.8\,\text{A}} \\ &\approx 82.143\,\Omega \end{aligned}$$

3. Relate the resistances to find the steady temperature ($T_2$):

Using the formula $R_2 = R_1 [1 + \alpha(T_2 - T_1)]$:

$$\begin{aligned} \frac{R_2}{R_1} - 1 &= \alpha(T_2 - T_1) \\ T_2 - T_1 &= \frac{1}{\alpha} \left( \frac{R_2}{R_1} - 1 \right) \\ T_2 - T_1 &= \frac{1}{\alpha} \left( \frac{V/I_2}{V/I_1} - 1 \right) \\ T_2 - T_1 &= \frac{1}{\alpha} \left( \frac{I_1}{I_2} - 1 \right) \end{aligned}$$

4. Substitute the numerical values:

$$\begin{aligned} T_2 - 27.0 &= \frac{1}{1.70 \times 10^{-4}} \left( \frac{3.2}{2.8} - 1 \right) \\ T_2 - 27.0 &= \frac{1}{1.70 \times 10^{-4}} \left( \frac{0.4}{2.8} \right) \\ T_2 - 27.0 &= \frac{1}{1.70 \times 10^{-4}} \left( \frac{1}{7} \right) \\ T_2 - 27.0 &\approx 840.336\,^{\circ}\text{C} \\ T_2 &\approx 840.336 + 27.0 \\ T_2 &\approx 867.336\,^{\circ}\text{C} \end{aligned}$$

Rounding to a reasonable number of significant figures based on the input data:

$$T_2 \approx 867\,^{\circ}\text{C}$$

The steady temperature of the heating element is approximately $867\,^{\circ}\text{C}$.

$$867.3\,^{\circ}\text{C}$$