What was the temperature of the object?

  • Thread starter Thread starter Math10
  • Start date Start date
  • Tags Tags
    Temperature
Math10
Messages
301
Reaction score
0

Homework Statement


An object is placed in a room where the temperature is 20 degrees Celsius. The temperature of the object drops by 5 degrees Celsius in 4 minutes and by 7 degrees Celsius in 8 minutes. What was the temperature of the object when it was initially placed in the room?

Homework Equations


Newton's law of cooling: T(t)=Ta+(To-Ta)e^(-kt)

The Attempt at a Solution


T(t)=Ta+(To-Ta)e^(-kt)
I know that I need to solve for To, which is the initial temperature but I don't know what to do.
 
Physics news on Phys.org
Rewrite Newton in terms of delta T and see if that gives you any ideas.
 
Math10 said:

Homework Statement


An object is placed in a room where the temperature is 20 degrees Celsius. The temperature of the object drops by 5 degrees Celsius in 4 minutes and by 7 degrees Celsius in 8 minutes. What was the temperature of the object when it was initially placed in the room?

Homework Equations


Newton's law of cooling: T(t)=Ta+(To-Ta)e^(-kt)

The Attempt at a Solution


T(t)=Ta+(To-Ta)e^(-kt)
I know that I need to solve for To, which is the initial temperature but I don't know what to do.
When t= 4 min, T(4)= Ta+ (To- Ta)e^(-4k)= 15. When t= 8 min, T(8)= Ta+ (To- Ta)e^(-8k)= 13.
What do you get if you subtract e^(-8k)T(4)- e^(-4k)T(8)?

You are asked to find T(0)= To.
 
  • Like
Likes Math10
I got it! Thanks everyone for the help!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top