Can someone help me with this math question ?

Question

A frozen steak has a temperature of 28 f. It is placed in a room with a constant temp of 70 f. After 10 mins, the temp of the steak has risen to 35 , what will the temp of the steak be after 30 mins?... There's more, but if I can figure out this part... I can do the rest

I thought you would do Newton's law of cooling, but it doesn't work... so I got confused.

Please help me understand how to get a formula for this.

all those answers are different... I'm still confused.

Answer 1

Hello,

Let's look at this as an interest problem.

a = p(1+r)^t except we will us f = 28(1 + i)^t where f is the temp in far degrees , i is the amount it increases per minute and t is the time. So we have 35 = 28(1 + i)^10 then 35/28 = (1+i)^10 so 5/4 = (1 + i) ^10 now take the 10 root of both sides giving us

(1.25)^(1/10) = 1 + i so i = (1.25)^(1/10) -1 = i then

1.0226 -1 = i so i = .0226

Now putting it back into the equation using 30 for the time we have

f = 28*(1 + 0.0226)^30 gives us 28*1.0692 = 54.69 degrees f


Hope This Helps!!

Answer 2

You are kind of right – in reverse!
This is a ‘ –Ae^-kt’ type with an asymptote of 70
So y - 70 – Ae^kt where y is the temp
So y – 70 = -Ae^-kt

For temp = 28, y – 70 at t = 0
is -42 = -A as e^(-k x0) = 1
So A = 42
At y = 35 and t = 10 35 – 70 = - 42 e^-kt
-35 = -42 e^-kt
0.8333 = e^-kt
Take lns and you are away!

Answer 3

Using the available data:
28 to 35 degrees f - 7 degrees/10 minutes = 0.7 degrees / min
0.7 degrees/min * 30 minutes = 21 degrees f
35 + 21 = 56 degrees f

Answer 4

without any further information, boundary conditions, overall heat transfer coefficient, etc., just assume linearity
7 F change in 10 mins, so 21 F change in 30 minutes, gives 59F
make sure you state that you are assuming a linear correlation.