Question

A thermometer is taken from a room where the temperature is 20°C to the outdoors, where the temperature is 3°C. After one minute the thermometer reads 11°C. (Round your answers to one decimal place.)

(a) What will the reading on the thermometer be after one more minute?

(b) When will the thermometer read 4°C?

Answer 1

Answer:

A. T=7.12°C; B. t=3.32min

Step-by-step explanation:

In order to find the answer we can use the Newton's law of cooling:

$y(t)=y0e^{kt}$ where:

y0=initial temperature - room temperature

k=coefficients

t=time

Because the outdoors temperature is 3°C, the initial temperature is 20°C, and because the thermometer reads 11°C after 1min we have:

$11=(20-3)e^{k*1}$

$ln(11/17)=k$

$-0.4353=k$

A. Temperature after an extra minute.

$y(2)=17e^{-0.4353*2}$

$y(2)=7.12$ so the answer is 7.12°C

B. Time in order to read 4°C

$4=17e^{-0.4353*t}$

$ln(4/17)/(-0.4353)=t}$

$3.32min=t$

Answer 2

(a) Reading on the thermometer after one more minute is 6.763°.

(b) After 3.78min the thermometer read 4°C.

What is newton's law of cooling?

According to newton's law of cooling

$T(t) = Ts + (To-Ts)e^(kt)$

Where

T(t) = temperature at time t

Ts =  temperature of the surround

T₀ = temperature of object at t=0

k = constant

It is given that

T₀ = 20°

Ts = 3°

After 1 minute  T₀ = 11°

$11 = 3 + (20-3)e^(k*1) \\\\k = -0.754$

Temperature after one more minute means temperature after 2 minutes

$T_{2} = 3 + (20-3)e^(-0.754*2)\\\\T_{2} = 6.763degree$

Time at which thermometer read 4°

$4 = 3 + (20-3)e^(-0.754t)\\t = 3.78min$

Therefore, (a) Reading on the thermometer after one more minute is 6.763°.

(b) After 3.78min the thermometer read 4°C.

To get more about Newton's law of cooling visit:

https://brainly.com/question/13748261