Athermometer reading 100◦f is placed in a medium having a constant temperature of 70◦f. after 6 min, the thermometer reads 80◦f.
what is the reading after 20 min?

Answers

b is 38 degrees.

a is 52 degrees.

this is because if b is 38 and a is 14 degrees more than b than 38+14=52 and 52+38=90 degrees. hope this .

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

70.00143

Step-by-step explanation:

Given that a thermometer reading 100◦F is placed in a medium having a constant temperature of 70◦F. After 6 min, the thermometer reads 80◦F.

We know that Newton law of cooling has formula as

T(t) = Ts +(T_0 - T_s)e^{-kt.}

where Ts is room temperature and To is initial temperature.

Using the fact that when t =6, T (6) = 80

T0 = 100 and Ts =70 we can find k

80 = 70+(100-70)e^{-6k} \\e^{-6k}=0.33333\\k = 0.1831

Using this we find temperature when t=20

T(20) = 70+(100-70)e^{-20*0.1831}\\T(20) = 70+(100-70)e^{-3.6624}\\\\T(20) = 70.00143



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