A thermometer is taken from a room where the temperature is 20◦C to the outdoors, where the temperature is 5◦C. After one minute the t

Question

A thermometer is taken from a room where the temperature is 20◦C to the
outdoors, where the temperature is 5◦C. After one minute the thermometer reads 12◦C.
(a) What will the reading on the thermometer be after one more minute?
(b) When will the thermometer read 6◦C?

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Allison 3 weeks 2022-01-06T05:56:46+00:00 1 Answer 0 views 0

Answers ( )

    0
    2022-01-06T05:57:51+00:00

    Answer:

    a) T(2) = 8.265: at time t = 2 minutes the temperature will be 8.265 degress

    b) 6 = T(3.55): the temperature will be 6 degrees at time t = 3.55 minutes.

    Step-by-step explanation:

    When dealing with temperature changes, it’s best to work with Newton’s Law of Cooling.

    T(t) = T_s + Ce^{kt}

    here:

    T(t) : the temperature in the room.

    T_s : ambient (or outdoor) temperature (that always remains constant, in our case: T_s = 5 )

    C\,\text{and}\,k: are constants

    Our conditions are provided:

    1) T(0) = 20

    2) T(1) = 12

    using the first condition

    T(0) = 5 + Ce^{k(0)}\\20 = 5 + C(1)\\C = 15

    using the second condition:

    T(1) = 5 + Ce^{k(1)}\\12 = 5 + Ce^{k}\\e^k = \dfrac{7}{C}

    we can use our calculated value of C to find k

    e^k = \dfrac{7}{15}\\k = \ln{(\dfrac{7}{15})}\\k = -0.7621

    Finally we can put these constants back in the main equation:

    T(t) = T_s + Ce^{kt}

    T(t) = 5 + 15e^{-0.7621t} or T(t) = 5 + 15e^{\ln{(\frac{7}{15})t}

    a) Reading after one more minute?

    so it’s asking:

    T(2) = ?

    T(2) = 5 + 15e^{\ln{(\frac{7}{15})(2)}}\\T(2) = \dfrac{124}{15} \approx 8.267

    Hence, after one more minute the temperature of the room will be 8.267 degrees

    b) When will it be 6 degrees?

    T(t) = 6?

    6 = 5 + 15e^{-0.7621t}\\\text{and solve for $t$}\\\\\dfrac{6-5}{15}=e^{\ln{(\frac{7}{15})}t}\\\ln{\left(\dfrac{1}{15}\right)} = \ln{\left(\dfrac{7}{15}\right)t}\\\ln{\left(\dfrac{1}{15}\right)} \div \ln{\left(\dfrac{7}{15}\right)} = t \approx 3.55\\

    Hence at t = 3.55 minutes the temperature of the room will be 6 degrees.

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