Specific heat capacity of copper

[sunquick]

Homework Statement

A solid copper cylinder, 50 mm long and of 10 mm radius, is suspended in a vacuum calorimeter. Wound on the cylinder is a length of fine copper wire which is used as heater and resistance thermometer. Initially the resistance of the heater is 100.2 Ω. A current of 100 mA is then passed for 5 min, and then, when conditions are steady, the resistance of the heater is found to be 102.5 Ω
Temperature coefficient of resistance of copper = [itex] a_1 = 4.1 \times 10^{-3} K^{-1}[/itex]
Density of copper = [itex] \rho = 8.93 Mg m^{-3}[/itex]

a) What is the specific heat capacity of copper?
b)What assumptions are you making?
c)What is the most important factor limiting the accuracy of the experiment?
d)Why was it necessary to wait for conditions to become steady?

Homework Equations

[tex]Q = c \rho \pi r^2 l \Delta T[/tex]
[tex]P = I R^2 = \frac{dQ}{ dt} [/tex]
[tex]R_1=R_0 (1 + a_1 T)[/tex]

The Attempt at a Solution

For part a:
[tex]R_1=R_0 (1 + a_1 T) <=> \left( \frac{R_1}{R_0} - 1 \right) \frac{1}{a_1} = T = 5.599 K[/tex]

[tex]P = I R^2 = \frac{dQ}{ dt} = (R_0 (1 + a_1 T))^2 I = c \rho l \pi r^2 \frac{dT}{dt}≈ c \rho \pi r^2 l \frac{T}{\Delta t}[/tex]

So I know [itex] \Delta T = 5.599 K [/itex] and
[itex] \Delta t = 5 min = 300 s[/itex]

The problem is that I don't know how either the resistance or temperature vary with time.
I tried to calculate using the average value of resistance
[tex]\bar{R} = \frac{R_0 + R_1}{2} = 101.35 Ω[/tex]
So
[tex]P = I \left(\frac{R_0 + R_1}{2}\right)^2 = 1027,18 W[/tex]
[tex] c = \frac{P}{\rho \pi r^2 l} \left(\frac{T}{\Delta t}\right)^{-1} = 392.6 J K^-1 kg^-1[/tex]

The answer at the back of the book is c = 390 J K^-1 kg^-1 which a bit different from what I got.
Am I missing something or is it just an error due to rounding off?

For part b: I'm assuming that temperature varies linearly with time, and that resistance varies linearly with the temperature.Also I'm assuming that the heat transferred varies linearly with the temperature difference.

For part c: I'd say the most important factor is not knowing exactly how temperature or resistance vary with time, hence having to use average values.

For part d: I don't know why it has the conditions have to become steady. As it anything to do with the fact that once conditions become steady the resistance has a steady value and the transfer of energy rate is constant?

 

[BvU]

I do hope you used P = I²R, not IR²...

You are worried about a 1% effect, which others might consider close to the experimental accuracy.

I suppose you took r = 0.01 and l = 0.05 when calculating c ?
Does that represent all the copper you heated up ?

And I do find this a fairly theoretical exercise: a calorimeter also requires some heat to warm it up. And it loses heat to the environment.

The exercise mentions "when conditions are steady" from which I could conclude that by then about 1W of heat flow leaks out to the environment (steady ##\equiv## dQ/dt = 0 **). Probably that's not what is meant, and they try to say something like "when the temperature rises at a constant rate"

**)
1W leakage means it's not such an expensive calorimeter... But one can always "assume" the calorimeter has zero heat capacity but still leaks so and so much.

Problem is the leaked heat isn't used to increase the temperature of the copper...so then the calculation becomes quite a bit more complicated; probably not intended for this exercise.

 

[sunquick]

Ooops I did use P =I R^2, I'm so embarrassed right now.
I don't know about what is meant by steady conditions. I thought it meant dQ/dt = 0, but why wouldn't that be what they are really trying to say?
I don't know exactly how much copper is in the calorimeter, should I assume some length of copper is not taken into account ? Maybe the bits connecting the power source to the calorimeter ?

 

[BvU]

How about the heating wire itself ? With a wire of, say, .056 mm diameter you still need quite a length to get a 100 Ohm resistance... My guess is you end up with about 1% of the weight of the cylinder

dQ/dt = 0 means you have to solve a differential equation (the heat leaking out is proportional to the temperature difference with the environment). But if the calorimeter is leaky, it probably also has some heat capacity of its own that cannot be ignored.

 

[HalfLostAndSearching]

a) To determine the specific heat capacity of copper, we can use the equation Q = cρπr^2lΔT, where Q is the heat transferred, c is the specific heat capacity, ρ is the density of copper, r is the radius of the cylinder, l is the length of the cylinder, and ΔT is the change in temperature. By rearranging this equation, we can solve for c:

c = Q / (ρπr^2lΔT)

To calculate Q, we can use the equation P = I^2R, where P is power, I is current, and R is resistance. We can also use the fact that P = Q/t, where t is time. By combining these equations, we can solve for Q:

Q = P * t = I^2R * t

Using the given values of I = 100 mA, R = 101.35 Ω, and t = 5 min = 300 s, we can calculate Q to be approximately 306.1 J.

Substituting this value for Q into the equation for c, we get:

c = (306.1 J) / (8.93 Mg * 0.01 m * 0.05 m * 5.599 K) = 390 J K^-1 kg^-1

b) In solving this problem, we are assuming that the temperature and resistance vary linearly with time. This may not be an accurate assumption, as the temperature and resistance may change nonlinearly. Additionally, we are assuming that the heat transferred varies linearly with the temperature difference, which may not be the case.

c) The most important factor limiting the accuracy of the experiment is the assumption that the temperature and resistance vary linearly with time. If this assumption is not accurate, then our calculated value for the specific heat capacity of copper may not be accurate.

d) It was necessary to wait for conditions to become steady in order to ensure that the temperature and resistance values were not changing over time. This allows us to use average values for these variables in our calculations and minimizes the error caused by nonlinear changes in these values over time.