Question

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

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

Answer 1

Answer: (A)

$\frac{(200{)}^2}{R_0(1+\alpha \left(T-T_0\right)}$
$T\to$ temperature at ' $t$'
$T_0\to$ temperature at $t=300K$
$T-T_0=\frac{500-300}{30}(t)$
$T-T_0=\frac{200}{30}t$
$T-T_0=\frac{20t}{3}$
$\underset{0}{\overset{30}{\int }}\frac{(200{)}^2}{100\left(1+\alpha \frac{20t}{3}\right)}dt=\frac{200\times 200}{100}\underset{0}{\overset{30}{\int }}\frac{dt}{1+\frac{20\alpha }{3}t}$
$=\frac{400\times 3}{20\alpha }\ell n\left(\frac{\frac{1+20\alpha }{3}\times 30}{1}\right)$
$120=100(1+\alpha (200))$
$1+(200)\alpha =\frac{6}{5}$
$(200\alpha )=\frac{1}{5}$
$\alpha =\frac{1}{1000}$
$=60,000\ell n\left(\frac{6}{5}\right)$