How to use ∫ (integral) to calculate heat dissipation

[Mad-Scientist_In-Training]

Hi;
I am going to build a coffee cup warmer plate using resistors - as an experiment.

Instead of using a USB port from my computer I am going to use a 5V, 700mA cell phone charger.

Using Ohms Law I have calculated that I there will be 7.14\[\Omega\] of resistance so I have ordered 3 x 2.4\[\Omega\] 5W resistors (Yageo - through-hole) and also 3 x .12\[\Omega\] resistors (Yageo - through-hole) in case they are all at the bottom end of the 5% tolerance for each resistor ((3x2.4) + (3x.12) x.95) = 7.182\[\Omega\] ... just slightly larger than the 7.14\[\Omega\] required for the circuit.

I would like to try to determine ahead of time what will be the heat dissipated from the resistors so I can 1) learn how to estimate heating of circuit components, and 2) know if I will need to use a larger or smaller supply (and resistor values).

I have not done any calculus before and I am wondering if someone can assist me with a quick tutorial on how to use the ∫ (integral) symbol to calculate the heat that would be dissipated from resistors.

I have this forumla W= ∫ V(t) I(t) dt ... with a t2 at the top of the ∫ symbol and a t1 at the bottom of the ∫ symbol.

Here's an image of the formula:


The values that I am using are:
5V,
.7A,
7.14\[\Omega\]
100% duty cycle - always on.

Thanks for any assistance.

EDIT: For clarification, I would like to try to determine the temperature that the resistors will achieve.

 

[Tahmid]

When you have any two of voltage, current and resistance constant, you don't need to use calculus.

P = VI = (V^2)/R = (I^2)R = [(5 * 5)/7.14] W = 3.5W

You may need to use a power supply with larger capacity. 700mA is the maximum rating of the power supply and you're operating it at/near maximum ratings.

Hope this helps.
Tahmid.

 

[Mad-Scientist_In-Training]

Thanks for the reply.

What I am specifically looking to figure out is how power (W) through the circuit relates to the temperature that the resistors will generate as the purpose of this little experiment is to create a heating / warming plate for my coffee / tea cup.

It's a simple / somewhat mindless experiment but it helps me in starting to learn about electronics.

I have numerous power supplies around and can use a more powerful supply for the project - that's no prob.

 

[chuckey]

If the coffee cup did not loose heat then you would not have to build your heater, so you must find out at what rate the coffee cup looses heat. 1. Find out the typical volume of its contents (300 ml?). Then find out the temperature of the poured hot water (90 deg C?), work out when the coffee is too cold (50 degrees C?) and the time its takes to get to this temperature. Now you know you have a volume of water, which drops so many degrees so you can work out how many calories have been lost/ persecond, multiply by 4.2 and this should give you a figure of watts. Which I hope is less then 3.5, but I suspect is more like 15 W.
Frank