Physics Lab Question-HELP!!!!?

Question

aluminum conductor (resistance 50 ohms, mass 15g, specific heat capacity 0.215) at 20 deg Celcius is connected to a 10V power source for 3 minutes. To determine the final temp of the aluminum, what formula would I use? (This is an ALGEBRA based physics question). Resistance remains constant.

Thanks in advance!

Answer 1

Hehehe. If you're having trouble with an algebra based physics problem, wait until you have one that's based on differential equations ☺
OK. You don't use *one* formula, you use several.
10 volts across 50Ω is V²/R = 2 Watts = 2 Joules/sec.
after 3 minutes (180 seconds) there are 360 J spent heating the conductor (we assume that it doesn't lose any heat.) Since there are 4.186J in a calorie, this is the equivalent of adding 360/4.186 = 86 calories of thermal energy to the conductor. Since aluminum is .215 cal/gram-degree then 86/(.215*15) = 26.66 is the temperature rise (in degrees C) so the final temperature would be 46.66C.

You gotta learn how to use *all* of the tools in the box ☺


Doug

Answer 2

Resistance depends on temperature in this manner:

R = p(L/A)

where
L is the length of the resistor
A is the cross-sectional area of the resistor and
p is the resistivity

p can be further reduced to expose the temperature element:

p = p_0[a(T - T_0) +1)

where T is the final temperature
T_0 is the initial temperature (20 C in your case)
a is the specific heat capacity and
p_0 is the resistivity coefficient (specific to aluminum in your case).

I don't see that p_0 is given in your question, so you may have to look it up. But Algebraically solve for T by use of that equation and you should be fine.

Answer 3

hi :-) As you boost the voltage the present additionally will boost inflicting the filament interior the easy bulb to get warm, the present starts off to even out by way of fact the filament has already replace into warm enought and can't get any warmer, the resistance additionally boost by way of fact once you employ the formula V=IR to get R we use R= V/I and if the voltage remains increasing by using a huge volume however the present isn't then you definately are in actuality getting a extra physically powerful resistance by way of fact the the present could extra or much less stay a similar yet you're dividing a bigger voltage with a similar contemporary so it is larger, it is a similar for the smaller contemporary, the resistance and voltage is very proportional by way of fact as you boost V, R has to boost to maintain the desirable contemporary for the time of the lighbulb, I=V/R if the voltage is growing to be and the resistance is aswell then i does not replace by way of fact the two V and R are growing to be :-) in case you wanna understand the rest e mail me ;-) i'm from the united kingdom aswell :-)

Answer 4

P= I squared x R = (V/R) squared x R = V squared / R
E = 180 x V squared / R
E = mc delta theta
delta theta = 180 x V squared/ Rmc
Final temp = (180 x Vsquared/Rmc) + 20

P is power; E is energy used in 3 min (180 s)
I is current ;V is pd; R is resistance; m is mass; c is sp heat cap; delta theta is change in temp