Q.At room temperature ($27.0\ ^\circ\text{C}$) the resistance of a heating element is $100\ \Omega$. What is the temperature of the element if the resistance is found to be $117\ \Omega$, given that the temperature coefficient of the material of the resistor is $1.70 \times 10^{-4}\ ^\circ\text{C}^{-1}$.

Concept: Temperature Dependence of Resistance — resistance changes linearly with temperature for most conductors over a moderate range.

Reasoning:

1. The relation is $R_T = R_0 (1 + \alpha \Delta T)$, where $R_0$ is resistance at reference temperature $T_0$, $\alpha$ is the temperature coefficient, and $\Delta T = T - T_0$.

2. Here $R_0 = 100\ \Omega$ at $T_0 = 27.0\ ^\circ\text{C}$, $R_T = 117\ \Omega$, and $\alpha = 1.70 \times 10^{-4}\ ^\circ\text{C}^{-1}$.

3. Rearranging: $\Delta T = \frac{R_T - R_0}{\alpha R_0} = \frac{117 - 100}{(1.70 \times 10^{-4})(100)} = \frac{17}{0.0170} = 1000\ ^\circ\text{C}$.

4. So $T = T_0 + \Delta T = 27.0 + 1000 = 1027\ ^\circ\text{C}$.

The key idea here is that for most metallic conductors, resistance increases linearly with temperature over a wide range. This is captured by the formula $R_T = R_0(1 + \alpha \Delta T)$, where $\alpha$ is the temperature coefficient of resistance. The problem gives us a reference resistance at room temperature and asks us to find the temperature at which the resistance rises to a new value.

Let’s work through it step by step.

1. Identify the known quantities.

   • Reference temperature, $T_0 = 27.0\ ^\circ\text{C}$
   • Resistance at $T_0$, $R_0 = 100\ \Omega$
   • Resistance at unknown temperature $T$, $R_T = 117\ \Omega$
   • Temperature coefficient, $\alpha = 1.70 \times 10^{-4}\ ^\circ\text{C}^{-1}$

2. Write the relation between resistance and temperature.

   The standard formula is:

This assumes $\alpha$ is constant over the temperature range — a reasonable approximation here.

3. Substitute the known values and solve for $T$.

Divide both sides by 100:

Subtract 1 from both sides:

4. Isolate $(T - 27)$.

Simplify the fraction:

5. Find the final temperature.

Method: Temperature Coefficient of Resistance Formula

This method uses the linear approximation for how resistance changes with temperature for most conductors.

Steps

1. Recall the formula The resistance at temperature $T$ is related to the resistance at a reference temperature $T_0$ by:

where:

• $R_T$ = resistance at temperature $T$ (unknown)
• $R_0$ = resistance at reference temperature $T_0$
• $\alpha$ = temperature coefficient of resistance

2. Identify given values

   • $R_0 = 100\ \Omega$ at $T_0 = 27.0\ ^\circ\text{C}$
   • $R_T = 117\ \Omega$
   • $\alpha = 1.70 \times 10^{-4}\ ^\circ\text{C}^{-1}$

3. Substitute into the formula

4. Solve for $T$
   • Divide both sides by 100:

• Subtract 1:

• Divide by $\alpha$:

• Add 27.0:

5. Final answer The temperature of the element is $1027\ ^\circ\text{C}$.

Key insight: The small value of $\alpha$ means resistance changes slowly with temperature — a 17% increase in resistance corresponds to a large temperature rise of $1000\ ^\circ\text{C}$ above room temperature.

Here are the common mistakes students make on this problem and how to avoid each one.

1. Using the wrong formula

Mistake:

Students often use the linear approximation formula incorrectly, or confuse it with the formula for resistivity change:

Some mistakenly write:

Why it’s wrong:

The formula uses the change in temperature ($\Delta T$), not the final temperature itself.

How to avoid:

Always write the formula as:

where $T_0$ is the reference temperature (here $27.0\ ^\circ\text{C}$) and $T$ is the unknown final temperature.

2. Forgetting to convert Celsius to Kelvin (unnecessary here)

Mistake:

Some students convert $27.0\ ^\circ\text{C}$ to $300.15\ \text{K}$ and then try to use the formula.

Why it’s wrong:

The temperature coefficient $\alpha$ is given in $^\circ\text{C}^{-1}$. Since a change of $1\ ^\circ\text{C}$ equals a change of $1\ \text{K}$, the formula works identically in Celsius. Converting to Kelvin adds extra steps and risk of error.

How to avoid:

Stick to Celsius when $\alpha$ is given in $^\circ\text{C}^{-1}$. Only convert if the problem explicitly asks for Kelvin.

3. Misidentifying $R_0$ and $R_T$

Mistake:

Plugging $R_0 = 117\ \Omega$ and $R_T = 100\ \Omega$ (swapping them).

Why it’s wrong:

$R_0$ is the resistance at the reference temperature ($27.0\ ^\circ\text{C}$), which is $100\ \Omega$. $R_T$ is the resistance at the unknown higher temperature, which is $117\ \Omega$.

How to avoid:

Label clearly:

• $T_0 = 27.0\ ^\circ\text{C}$, $R_0 = 100\ \Omega$
• $T = ?$, $R_T = 117\ \Omega$

4. Arithmetic errors in solving for $T$

Mistake:

After substituting, students often make sign errors or misplace decimals when isolating $T$.

Correct steps:

Divide both sides by 100:

Subtract 1:

Divide by $1.70 \times 10^{-4}$:

So:

How to avoid:

Write each step clearly. Double-check the division: $0.17 \div (1.70 \times 10^{-4}) = 0.17 \div 0.00017 = 1000$.

5. Forgetting to add back the reference temperature

Mistake:

Stopping at $\Delta T = 1000\ ^\circ\text{C}$ and writing the answer as $1000\ ^\circ\text{C}$.

Why it’s wrong:

The question asks for the temperature of the element, not the change.

How to avoid:

Always finish with:

Quick checklist to avoid all mistakes

Final correct answer: $\boxed{1027\ ^\circ\text{C}}$