# Newton's Law of Cooling, find room temperature

• hulgy
In summary: Celsius!In summary, the conversation discusses using Newton's law of cooling to find the room temperature of a pan of warm water after it has been placed in a room for 20 minutes and its temperature has decreased from 47 degrees Celsius to 34 degrees Celsius. By setting up two equations and substituting M=e10k, the room temperature is found to be -2.111 degrees Celsius.
hulgy

## Homework Statement

Suppose that the temperature of a pan of warm water obeys Newton's law of cooling. The water (47 degrees Celsius) was put in a room and 10 minutes later the water's temperature was 40 degrees Celsius. After another 10 minutes, the temperature of the water was 34 degrees Celsius. Find the room's temperature.

## Homework Equations

(H-Hs)=(H0-Hs)ekt
dH/dt = -k(H-Hs)

Hs is the room's temperature
H0 is the initial temperature of the water
H is the temperature of the water at time t

## The Attempt at a Solution

I'm rather at a loss at how to do this problem. I've tried setting up a system of equations and tried finding the room temperature, but that doesn't work out. So if algebra won't do it, then I'm assuming something calculus-y should. I'm just at a lost as what that is.

Last edited:
You have the temperatures at 10 minutes and at 20 minutes. The exponential terms are e10k and e20k which is the square of e10k . Plug in the temperature data and use that the second equation is just the square of the first one.

ehild

It doesn't work, it only leads to a polynomial that can't be factored.

If it helps at all Newton's law of cooling is dH/dt = -k(H-Hs), the equation above is derived from it.

It didn't lead to an unfactorable polynomial when I tried it. . . I ended up substituting M=e10k, which made it easier to solve (or less intimidating, anyways).

It works, just try.

Let be z=e10k. Then e20k=z^2.

You have two equations:

40-Hs=(47-Hs)z
34-Hs=(47-Hs)z^2

What about squaring the first equation and dividing by the second one?

ehild

OK..

40-Hs=(47-Hs)z
34-Hs=(47-Hs)z^2

I square the first equation
(40-Hs)^2=(47-Hs)^2*z^2

and divide by the second, so I get:
(40-Hs)^2/(34-Hs) = (47-Hs)

...FFFFFUUUUUUUUUUUUUUUUUUUUU

I was doing this all along, but I was writing some of my signs wrong... so yeah. Anyway thanks for the help, answer is -2.

## 1. What is Newton's Law of Cooling?

Newton's Law of Cooling is a mathematical equation that describes the rate at which an object cools down in a surrounding environment with a different temperature.

## 2. How does Newton's Law of Cooling work?

According to Newton's Law of Cooling, the rate at which an object cools is directly proportional to the temperature difference between the object and its surrounding environment. This means that the larger the temperature difference, the faster the object will cool down.

## 3. What is the formula for Newton's Law of Cooling?

The formula for Newton's Law of Cooling is T(t) = Ta + (To - Ta)e^(-kt), where T(t) is the temperature of the object at time t, Ta is the temperature of the surrounding environment, To is the initial temperature of the object, and k is the cooling constant.

## 4. How can Newton's Law of Cooling be used to find room temperature?

If we know the initial temperature of an object and the surrounding environment, we can use Newton's Law of Cooling to calculate the cooling constant, k. Once we have the value of k, we can plug it into the formula and solve for the temperature of the object at any given time. By setting t to a large value, we can estimate the room temperature as the final temperature of the object.

## 5. What are some real-life applications of Newton's Law of Cooling?

Newton's Law of Cooling has various practical applications, such as predicting the cooling of hot beverages, determining the temperature of food in a refrigerator, and estimating the time of death in forensics. It is also used in engineering and physics to study heat transfer and thermodynamics.

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