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  • The resistance of an electrical toaster has a temperature dependence given by R(T)=R0 [1 + α(T−T0)] in its range of operation.

The resistance of an electrical toaster has a temperature dependence given by

R(T)=R0 [1 + α(T−T0)] in its range of operation.  At T0=300 K, R=100 and at T=500 K,

R=120 .  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 :

Option: 1

400\, \l n\frac{1.5}{1.3}J


Option: 2

200\, \l n\frac{2}{3}J


Option: 3

400 \l n\frac{5}{6}J


Option: 4

300\, J


Answers (1)

best_answer

\\ Given: R(T)=R_{o}\left(1+\alpha\left(T-T_{o}\right)\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(t)=300+(20t/3)

\\ By \ Joule's \ Law, \ heat \ dissipated \ in \ a \ resistor \ is \ given \ by: \\ \\ W=\int_{0}^{30} \frac{V^{2}}{R} d t =\int_{0}^{30} \frac{V^{2}}{R_{o}\left(1+\alpha\left(T-T_{o}\right)\right)} d t =\frac{V^{2}}{R_{o}} \int_{0}^{30} \frac{1}{(1+(20 \alpha t / 3))} d t \\ \\ Solving, W=400 \ln (6 / 5) \\ \text{Work done on resistor} =-W=400 \ln (5 / 6) J

Posted by

sudhir.kumar

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