The voltage of a certain thermocouple as a function of the temperature is given by E = 2.800T + 0.006#(T∧2)#. If the temperature is increasing at the rate of 1.00°C/min,how fast is the voltage increasing when T=100°C?

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Answer 1 · 2018-01-25T10:59:38

The voltage is increasing at the rate of #=4.0Vmn^-1#

The relation is

#E=2.8T+0.006T^2#

Differentiating with respect to #t#

#(dE)/dt=2.8(dT)/dt+0.006*2T*(dT)/dt#

#(dE)/dt=2.8(dT)/dt+0.012T(dT)/dt#

Here, #(dT)/dt=+1.00^@Cmn^-1#

and

#T=100^@C#

Therefore,

#(dE)/dt=2.8*1+0.012*100*1#

#=2.8+1.2#

#=4.0Vmn^-1#