A Nichrome heater dissipates 500 Wwhen the applied potential difference is 110 Vand the wire temperature is$\mathbf{800}\mathbf{°}\mathit{C}$. What would be the dissipation rate if the wire temperature were held at$\mathbf{200}\mathbf{°}\mathit{C}$by immersing the wire in a bath of cooling oil? The applied potential difference remains the same, and for Nichrome at$\mathbf{800}\mathbf{°}\mathit{C}$isrole="math" localid="1661414566428" $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}{\mathit{k}}^{\mathbf{-}\mathbf{1}}$.

The given data can be listed below as:

• Power dissipated by the heater is,$P_H=500W$ .
• Applied potential difference is,$V=110V$ .
• Higher temperature of the resistance is,$T_H=\left(800°C+273\right)K=1073k$ .
• Lower temperature of the resistance is,$T_L=\left(200°C+273\right)K=473k$

Using the concept of resistance change due to a change in the temperature value, we can get the resistance at the given lower temperature. Now, using this resistance value in the power dissipation formula, we can get the rate of dissipated energy.

Formulae:

The resistance value after linear expansion of a material is,

$R=R_0+R_0\alpha ∆T$ … (i)

The electric power generated during dissipation is,

$P=\frac{V_2}{R}$ … (ii)

Let$R_H$ be the resistance value at temperature of$800°C$ .

Let$R_L$ be the resistance value at temperature of$200°C$ .

So, for the resistance value at temperature, we can use the formula of equation (i) as follows:

$R_L=R_H\left(1+\alpha \left(T_L-T_H\right)\right)$

Now, using this above value and equation (i) for the resistance in equation (ii), we can get the value of the dissipated energy rate or the power at temperature as follows:

$$\begin{gathered} P_L=\frac{R_H\left(P_H\right)}{R_H\left(1+\alpha \left(T_L-T_H\right)\right)} \\ =\frac{P_H}{1+\alpha \left(T_L-T_H\right)} \end{gathered}$$

Here,is the thermal expansion coefficient of nichrome whose value is$4\times {10}^{-4}/k$ .

Substitute the values in the above equation.

$$\begin{gathered} P_L=\frac{500W}{1+\left(4\times {10}^{-4}/K\right)\left(473-1073\right)K} \\ \approx 660W \end{gathered}$$

Hence, the required value of the rate of dissipation is 660W .