Question

i

Answer 1

Hints:
(a) The rate of temperature change is proportional to the temperature difference. Define some arbitrary constant of proportionality. For example, if \(T\) is the pizza temperature, \(T_0\) is the refrigerator temperature, and \(k\) is some arbitrary constant, you could write the following.\[\frac{dT}{dt}=k(T_0-T)\](b) Solve the differential equation in part (a).
(c) Plug in the initial conditions to solve for constants and evaluate at \(t=0\).
(d) Give this one a bit of thought. You don't need any math to do this one.

Answer 2

Let T = temperature of pizza
\[Let\ T _{i}=initial\ temperature\ of\ pizza\]
Let k = decay constant
Let t = elapsed time in minutes
\[\frac{dT}{dt}=-kT _{i}e ^{-kt}\]
\[T=T _{i}e ^{-kt}.................(1)\]
The given points on the decay curve can be used to form a pair of simultaneous equations:
\[40=T _{i}e ^{-20k}............(2)\]
\[20=T _{i}e ^{-40k}............(3)\]
Now you need to solve to find k and Ti

Answer 3

Hint: To solve equations (2) and (3) first take natural logs of both sides of both equations. Then subtract the resulting equations. Can you do that?

Answer 4

can you pls explain how u got eqn 1

Answer 5

Equation (1) uses the equation for exponential decay that applies to radioactive decay, disharge of electrical capacitors and so on.

Answer 6

ok, but shouldn't we be using newton's law of cooling

Answer 7

Newton's law of cooling is basically exponential decay. I have calculated the initial temperature of the pizza. If you have answers available do you want to check my answer?

Answer 8

no i do not have the answers

Answer 9

Understood. My calculated value of Ti came to a nice round figure which is reassuring :)

Answer 10

my k is 0.0347

Answer 11

\[k=\frac{\ln 40-\ln 20}{20}=0.034657359\]

Answer 12

yes that is what i got and Ti is 80.07

Answer 13

Good. My calculator gives Ti = 79.99999994

Answer 14

i know ti is the answer for c but wat is the solution for b

Answer 15

The solution for b is found by substituting the values of Ti and k into equation (1)

Answer 16

\[T=80e ^{-0.03466t}\]

Answer 17

To solve d) we put T = 0 in the solution for b)
\[0=80e ^{-0.03466t}\]
According to my calculator e^-230 = 0
Therefore -0.03466t = -230
\[t=\frac{-230}{-0.03466}=?\]