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2 years ago

After heating up in a teapot, a cup of hot water is poured at a temperature of

Mathematics

1 answer:

xxTIMURxx [149]2 years ago

3 0

Answer:

k ≈ 0.060
T(4) ≈ 178 °F

Step-by-step explanation:

The desired formula parameters for Newton's Law of Cooling can be found from the given data. Then the completed formula can be used to find the temperature at the specified time.

<h3>Given:</h3>

$T(t)=T_a+(T_0-T_a)e^{-kt}\\\\T_a=68,\ T_0=208,\ (t,T)=(3,185)$

T(4)

<h3>Solution:</h3>

Filling in the given numbers, we have ...

185 = 68 +(208 -68)e^(-k·3)

117/140 = e^(-3k) . . . . . subtract 68, divide by 140

ln(117/140) = -3k . . . . . . take natural logarithms

k = ln(117/140)/-3 ≈ 0.060

The temperature after 4 minutes is about ...

T(4) = 68 +140e^(-0.060·4) ≈ 68 +140·0.787186

T(4) ≈ 178.205

After 4 minutes, the final temperature is about 178 °F.

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