Question

The resistance of an electrical toaster has a temperature dependence given by $R\left(T\right)=R_0[1+\alpha (T-T_0\left)\right]$ in its range of operation. At $T_0=300K,R_0=100\Omega$ and at $T=500K,R=120\Omega$. The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate from 300 to 500 K in 30 s. The total work done in raising the temperature is :

• A. $400ln\frac{5}{6}J$
• B. $200ln\frac{2}{3}J$
• C. $300J$
• D. $400ln\frac{1.5}{1.3}J$

Answer 1

The correct option is A $400ln\frac{5}{6}J$

Given: $R\left(T\right)=R_o(1+\alpha (T-T_o\left)\right)$

Applying boundary conditions,

$120=100(1+200\alpha )$

$\alpha ={10}^{-3}{K}^{-1}$

It is given that temperature increases at a constant rate from 300K to 500K in 30s.

Hence, $T\left(t\right)=300+20t/3$

By Joule's Law, heat dissipated in a resistor is given by:

$W={\int }_0^{30}\frac{V^2}{R}dt$

$={\int }_0^{30}\frac{V^2}{R_o(1+\alpha (T-T_o\left)\right)}dt$

$=\frac{V^2}{R_o}{\int }_0^{30}\frac{1}{(1+20\alpha t/3)}dt$

Solving, $W=400ln\left(6/5\right)$

Work done on resistor $=-W=400ln\left(5/6\right)J$