Question

A thermometer is taken from a room where the temperature is 20◦C to the

Answer 1

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.