Question

a steel ball is placed at time t=0 in a medium which is maintained at a temperature of 30 C. heat flows so rapidly within the ball that the temperature is essentially the same at all points of theball. at the end of 3 minutes, the temperature of the ball reduced tp 70 C. after 15 minutes the temperature of the ball becomes 42 C. find the initial temperature of the ball

Answer 1

Given:

T_m = 30 C

@ t = 0
T = ?

@ t = 3
T = 70

@ t = 15
T = 42

Answer 2

\[\ln (\frac{42 - 30}{70-30} )= k(15-3)\]
\[\ln (0.3) = 12k\]
\[k= -0.100\]

Answer 3

\[\ln (\frac{T - 30}{42 - 30}) = -0.100 (3)\]
\[\ln (\frac{T - 30}{12}) = -0.301\]
\[\frac{T-30}{12} = e^{-3.01}\]
\[T = 30 + 12e^{-3.01}\]
\[T = 38.89 C\]
what did i do wrong

Answer 4

−0.100(0-15) not −0.100(3)

Answer 5

first line of the last reply

Answer 6

\[\ln(\frac{T-30}{42-30})=-0.1(0-15)\]

Answer 7

oh lol of course thanks

Answer 8

yw

Answer 9

so \[\ln (\frac{T-30}{12}) = 1.5\]

\[T = 30 + 12e^{1.5}\]
\[T = 84^o C \checkmark\]