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 $T_0 = 300 \,K, R=100 \Omega$ and at $T=500 \, K, 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. $400 \, ln\, \frac{1.5}{1.3} \,J$
• B. $200 \, ln\, \frac{2}{3}\, J$
• C. $400 \, ln \,\frac{5}{6}\, J$
• D. $300 \, ln \,\frac{5}{6}\, J$

Answer 1

Answer: (A)

$\frac{(200)^{2}}{ R _{0}\left(1+\alpha\left( T - T _{0}\right)\right.}$
$T \rightarrow$ temperature at ' $t$'
$T _{0} \rightarrow$ temperature at $t =300\, K$
$T - T _{0}=\frac{500-300}{30} (t)$
$T - T _{0}=\frac{200}{30} t$
$T - T _{0}=\frac{20 t }{3}$
$\int\limits_{0}^{30} \frac{(200)^{2}}{100\left(1+\alpha \frac{20 t}{3}\right)} d t=\frac{200 \times 200}{100} \int\limits_{0}^{30} \frac{d t}{1+\frac{20 \alpha}{3} t}$
$=\frac{400 \times 3}{20 \alpha} \ell n \left(\frac{\frac{1+20 \alpha}{3} \times 30}{1}\right)$
$120=100(1+\alpha(200))$
$1+(200) \alpha=\frac{6}{5}$
$(200 \alpha)=\frac{1}{5}$
$\alpha=\frac{1}{1000}$
$=60,000 \,\ell n \left(\frac{6}{5}\right)$