A thermometer is taken from a room where the temperature is 20°C to the outdoors, where the temperature is 6°C. After one minute the thermometer reads 13°C. (Round your answers to one decimal place.)

Question

a) what will the reading the thermometer be after one minute?

b) when will the thermometer read 7°C?

Answer 1

dT/dt = - k(T - 6); ; dT/(T - 6) = - kdt; ln(T - 6) = - kt + lnC; T - 6 = Ce^{- kt}; If t = 0 then T = 20°. C = 14°;

T = 6 + 14e^{- kt;}; 13 = 6 + 14e^-k; k = - ln(7/14) = ln2;

So, T = 6 + 14e^{- tln2} = 6 + 14/2^t;

a) t = 1 min then T = 6 + 14/2 = 13°C

b) T = 7°; 7 = 6 + 14/2^t; 2^t = 14; t = ln14/ln2 = 3.81 min

Answer 2

Let T = temperature reading on the thermometer t minutes after being taken outside.

By Newton's Law of Cooling, dT/dt = k(T - 6). T(0) = 20 and T(1) = 13

(1 / (T - 6))dT = kdt

Integrate to get ln (T - 6) = kt + C

T - 6 = e^kt+C

T(t) = e^Ce^kt + 6. Relabel e^C as C, So T(t) = Ce^kt + 6

Since T(0) = 20, we have 20 = C + 6. So, C = 14.

T(t) = 14e^kt + 6

T(1) = 14e^k + 6 = 13. So, e^k = 0.5. k = ln(0.5) = -0.693.

T(t) = 14e^-0.693t + 6

If T(t) = 7, then 14e^-0.693t + 6 = 7

e^-0.693t = 1/14

-0.693t = ln(1/14)

t = 3.8 minutes