A metal ball at room temperature 20° C is dropped into a container of boiling water (100° C).?
given that the temperature of the ball increases 2° in 2 seconds, find:
(a) The temperature of the ball after 6 seconds in the boiling water.
(b) How long it will take for the temperature of the ball to reach 90° C.
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The temperature difference decays exponentially;
100 - T = ce^(-kt)
Given that T = 20 at t = 0, the "c" is 80.
100 - T = 80e^(-kt)
Given that T = 22 at t = 2s, the "k" can be found:
78 = 80 e^(-k(2 seconds))
ln(80/78) = k*(2 seconds)
k = 0.012659 s^(-1)
(a) Now put in t = 6 seconds:
100 - T = 80 e^(-0.07595) = 0.9269*80
T = 25.85
(b) Now put in T - 100 = 10;
10/80 = e^(-0.012659 t)
ln(8) = 0.012659 t
t = 164 seconds (2.7 minutes)