A calculator is required to obtain the final answer on this question. A solid metal sphere at room temperature 20oC is dropped into a container of boiling water (100oC). If the temperature of the sphere increases 10o in 9 seconds, find the temperature of the ball after 18 seconds in the boiling water. (Assume the sphere obeys Newton's Law of Cooling.)

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fichoh fichoh · Answer 1 · 2021-02-13T16:24:30+01:00

Answer:

38.71°C

Step-by-step explanation:

Given :

Initial temp, T0 = 20°C

Final temperature, T = 20 + 10 = 30°C

Time, t = 9 seconds

Surrounding temperature, Ts = 100°C

Newton's Law of cooling :

T = Ts + (T0 - Ts) * e^-kt

Obtain the value of k

30 = 100 + (20 - 100) * e^-9k

30 - 100 = - 80e^-9k

-70 = - 80e^-9k

-70 / - 80 = e^-9k

0.875 = e^-9k

Take the In of both sides

In(0.875) = - 9k

−0.133531 = - 9k

k = 0.133531 / 9

k = 0.0148

Hence,

t = 18

T = Ts + (T0 - Ts) * e^-kt

T = 100 + (20 - 100) * e^-0.0148(18)

T = 100 - 80 * e^-0.0148(18)

T = 100 - 80 * 0.7661326

T = 38.709392

T = 38.71°C