Figuring Out Time Required to Heat a Mass at Differing Temperatures

[Fancy Moses]

I have a fairly basic (I hope) question:

Given:

I'm heating, via convection, a constant sized spherical mass (Let's say m=1g) with a consistent surface area (Let's say SA = 1cm^2) at a constant temperature, Tc, (Let's say Tc = 60ºC) for a time, t = 60 minutes. Let us assume that the mass has an initial temperature (Ti) of 20ºC and after 60 minutes the temperature (Tf) of the mass will equal 50ºC. In a similar scenario (m=1g, SA=1cm^2, Ti=20ºC) how long will it take the mass to equal 50ºC if the new constant ambient temperature equals 100ºC?

Experiment 1:

m = 1g
SA = 1cm^2
Ti = 20ºC
Tc = 60ºC
Tf = 50ºC
t = 60 min

Experiment 2:

m = 1g
SA = 1cm^2
Ti = 20ºC
Tc = 100ºC
Tf = 50ºC
t = ?? min

How would I figure this out? Assuming any necessary chemical or physical properties can be found for the mass.

Any information would be greatly appreciated!

 

[dulrich]

You are going to want to use Newton's law of cooling.

http://en.wikipedia.org/wiki/Heat_transfer#Newton.27s_law_of_cooling

 

[Glen Bartusch]

Please do not post homework questions on this forum, as this forum is for the discusion of physics. Thank you.

 

[Fancy Moses]

This is not a homework question at all - it is a small part of a larger issue I am having testing different products in an environmental chamber. Please forgive me for formatting the question in a "homework-fashion". I thought it would be much easier to understand this way. I really hope this wasn't the reason for all the lack of responses...

 

[sardar]

I would approach this question by using the principles of thermodynamics and heat transfer. The rate at which the temperature of an object changes is determined by the heat transfer equation, which takes into account the mass, surface area, and temperature difference between the object and its surroundings. In this case, the only variable that changes between Experiment 1 and Experiment 2 is the ambient temperature (Tc), so we can use the following equation:

Q = mCΔT

Where Q is the heat transferred, m is the mass of the object, C is the specific heat capacity of the object, and ΔT is the change in temperature.

In Experiment 1, the heat transferred to the object can be calculated as:

Q = (1g)(1cm^2)(50ºC - 20ºC) = 30 calories

Using the specific heat capacity of water (since the mass is 1g and the object is assumed to have similar properties to water), we can calculate the time it takes for the object to reach 50ºC:

30 calories = (1g)(1 cal/gºC)(50ºC - 20ºC)t

t = 30 minutes

In Experiment 2, the heat transferred to the object will be the same (since the mass and surface area are constant), but the temperature difference will be greater (50ºC - 100ºC). Therefore, the time required for the object to reach 50ºC will be longer. Using the same equation as above, we can calculate the time as:

30 calories = (1g)(1 cal/gºC)(50ºC - 100ºC)t

t = 60 minutes

In conclusion, it will take 60 minutes for the object to reach 50ºC in Experiment 2, compared to 30 minutes in Experiment 1, due to the higher ambient temperature. Of course, this is a simplified scenario and in reality, there may be other factors at play, such as the efficiency of the heating source and any heat loss from the object. But this approach gives us a rough estimate of the time required in each experiment.