Newton's Law of Cooling Problem

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SUMMARY

The Newton's Law of Cooling problem involves a small metal bar initially at 20 degrees Celsius, placed in boiling water at 100 degrees Celsius. The bar's temperature increases by 2 degrees per second, reaching 90 degrees Celsius in 82.1 seconds and 98 degrees Celsius in 145.7 seconds. The differential equation governing the temperature change is given by \(\frac{dT}{dt}=k(T-T_m)\), where \(T_m\) is the surrounding temperature. The correct temperature function derived is \(T(t)=-80e^{t\ln(0.975)}+100\).

PREREQUISITES
  • Understanding of differential equations
  • Familiarity with Newton's Law of Cooling
  • Basic knowledge of temperature conversion and units
  • Concept of equilibrium temperature in thermodynamics
NEXT STEPS
  • Study the derivation of Newton's Law of Cooling in detail
  • Learn about solving first-order differential equations
  • Explore applications of temperature change in different materials
  • Investigate the effects of atmospheric pressure on boiling points
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Students studying physics or engineering, particularly those focusing on thermodynamics and differential equations, as well as educators teaching these concepts.

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Homework Statement



A small metal bar, whose initial temperature was 20 degrees C is dropped into a large container of boiling water. How long will it take the bar to reach 90 degrees C if it is known that its temperature increases 2 degrees in 1 second. How long will it take the bar to reach 98 degrees C?

(the back of the book gives 82.1 seconds and 145.7 seconds)

Homework Equations



\frac{dT}{dt}=k(T-Tm)

where T_m is the temperature of the surroundings.

The Attempt at a Solution



\int{\frac{1}{T-T_m}\frac{dT}{dt}dt=\int kdt

T(t)=Ce^{kt}+T_m

I'm given the points T(0)=20 and T(1)=T(0)+2=22

The problem is that this give two equations and three unknowns:

C+T_m=20

Ce^k+T_m=22

I'm given that the water is boiling, but that only gives a minimum temperature of 100 degrees celsius.
 
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Actually if the water is boiling you can be sure what the temperature is, as it will be a function of the atmospheric pressure only. Assuming the experiment is performed at the sea level, the temperature of the boiling water shall be 100 C.
 
Quiablo said:
Actually if the water is boiling you can be sure what the temperature is, as it will be a function of the atmospheric pressure only. Assuming the experiment is performed at the sea level, the temperature of the boiling water shall be 100 C.

Ok, you're right! I actually haven't taken chem yet. This is a differential equations application problem. Are you getting this from PV=nRT ?I guess if you have T=\frac{PV}{nR} I guess I can see how T is a function of pressure assuming constant volume.

I should have just tried using 100 C, because it solves the system and gives the correct function T(t)=-80e^{tln(.975)}+100 and t=82.1 gives 90 C.
 
No, no, PV = nRT is used only for (ideal) gases, and we are talking about water, a liquid that is. Indeed, the fact that water boils at 100 degrees C and no other temperature, assuming one atmospheric pressure, has to do with the fact that any substance changes its state (solid to liquid or liquid to gas or gas to liquid or liquid to solid) at a specific temperature.
 

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