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Question:

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=300K, R₀=100Ω and at T=500K, R=120Ω. 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 :

200ln23J

400ln1.51.3J

400ln56J

300J

Solution:

Given:R(T)=R₀(1+α(T−T₀))
Applying boundary conditions,120=100(1+200α)α=1/200 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=∫₃₀₀⁵₀₀ V²/R dt=∫₃₀₀⁵₀₀ V²/R₀(1+α(T−T₀)) dt=V²/R₀∫₃₀₀⁵₀₀ 1/(1+20αt/3) dt
Solving,
W=400 ln(6/5)
Work done on resistor=−W=400 ln(5/6) J