dq = λa T Direction of heat flow dq = λa T h T c

Size: px

Start display at page:

Download "dq = λa T Direction of heat flow dq = λa T h T c"

Melvyn Gordon
5 years ago
Views:

1 THERMAL PHYSIS LABORATORY: INVESTIGATION OF NEWTON S LAW OF OOLING Along a well-insulated bar (i.e., heat loss from the sides can be neglected), it is experimentally observed that the rate of heat flow through an area A is proportional to A and the local temperature gradient. The constant of proportionality is the thermal conductivity λ: dq = λa T (1) x where the minus sign is required because heat flows in the direction of decreasing temperature. In a steady state, when there are no further variations of the temperature with time, the temperature varies linearly with position along the bar: Th Tc Th Direction of heat flow T c Position along the bar, x 0 L In other words, T/ x is constant and equal to (T c T h )/L, sothateq.(1)above simplifies to the form you used in first year: dq = λa T h T c () L The rate at which a hot solid object surrounded by a fluid (e.g., air or water) cools is described well by a boundary-layer model, in which heat is transported from the object as described above through a thin layer of fluid that is at rest and in contact with the surface of the object. Outside this boundary layer of fluid, the heat is rapidly transported away from the object by currents moving in the fluid. Hot object, Temperature T b Temperature of Surroundings, T o Boundary layer, thickness b Surface area A The moving fluid outside the boundary layer acts to continuously remove the fluid warmed by heat conducted from the object, so that the temperature difference across the boundary layer can remain roughly constant. If the cooling is slow enough that quasi-steady-state conditions are established in the narrow boundary layer of fluid with thickness b, then the heat flow through the boundary layer will be dq = λa T T o b 1 (3)

2 where T is the temperature of the object and T o is the temperature of the surrounding fluid. A hot object whose rate of cooling is well described by this model is said to exhibit Newtonian cooling. AlossofheatenergydQ by the object results in a decrease in its temperature by dq = dt,where is the heat capacity of the object. We can use this relation to rewrite the heat flow through the boundary layer dt = λa T T o b dt = k(t T 0 ) (4) where k = λa/b is playing the role of the inverse of a characteristic cooling time, τ =1/k. We can rewrite the above equation more compactly in terms of the quantity Θ T T o (note that dθ/ = dt/, because T 0 is a constant): dθ = kθ (5) This is called Newton s law of cooling. This can easily be integrated to give dθ Θ = k dθ Θ = k ln Θ = kt +const (6) The constant of integration can be determined from the initial temperature difference Θ i = T i T o at time t =0: ln Θ i = k(0) + const ln Θ i = const ln Θ = kt +lnθ i Θ(t) = Θ i e kt (7) Thus, the initial temperature difference Θ i = T i T 0 decreases exponentially with time, with the characteristic decay time τ = 1/k. The cooling coefficient k can be determined experimentally from the slope of a linear fit to a plot of ln Θ as a function of time. If in addition we provide heat to the object in some way, there will be two competing processes affecting the temperature: the heating and the Newtonian cooling. If P (t) is the instantaneous heating power, the net infinitesimal change in temperature dt net will be dt net = dt heat + dt cool = dq heat k(t T o) = P (t) k(t T o) (8)

3 Integrating this equation and determining the constant of integration by taking t = 0 to be a time just before the heating starts, when the temperature of the object is equal to the background temperature, T = T o, yields dt net = 1 P (t) k (T (t) T o ) T (t) = T o + 1 P (t) k (T (t) T o ) (9) = T o +(ΔT ) heating (ΔT ) cooling In this laboratory, you will investigate the time dependence of the temperature of water that is heated and then allowed to cool. EXPERIMENTAL PROEDURE The apparatus consists of a calorimeter, stirrer, batteries, an ammeter, two thermometers, a balance and a timer. The heating is provided by a simple electrical circuit. The cooling is provided by conduction through a boundary layer of air surrounding the copper part of the calorimeter. 1. Measure the mass m of the empty calorimeter and stirrer, and the radius r and height h of the copper surface of the calorimeter, making a note of the uncertainties in these quantities.. Approximately half fill the calorimeter with water, ensuring the heating coil is covered. Determine the mass M of the water in the calorimeter together with its uncertainty. 3. Place a thermometer in the water and cover the calorimeter with its cork cover to minimise evaporation and heat conduction through the cover (the cork has a low heat capacity). 4. After allowing the apparatus to settle, read the initial temperature of the water and surrounding air (hopefully nearly the same). 5. With the timer switched on and zeroed, connect the heating circuit (ensure the two batteries are connected in series, i.e., positive to negative) and simultaneously start the timer. 6. Take a temperature and current measurement every minute for 30 minutes, stirring the water gently before each reading to help ensure the temperature of the water in the calorimeter is uniform. Note the temperature of the surroundings every 5 minutes using a second thermometer. You should have enough time between readings to enter your data into a spreadsheet as you go but make sure to make a written record of your data as well, to ensure you have a copy if you or someone else mistakenly deletes your computer file at some point. 7. Disconnect the heating circuit after 30 minutes and then continue to take temperature readings as before until the temperature has dropped to within a degree or so of the temperature of the surroundings. You may stop if this has not happened after an hour of cooling (i.e., 90 minutes in all). 3

4 DATA ANALYSIS You are free to use any software package you wish to do this data analysis. You will need to have your time and temperature readings entered as two columns of data, with the temperature of the surroundings T o entered in a third column (and if you wish, Θ = T T o entered in a fourth column). General Examination of the Data Make a plot of temperature vs. time for your entire dataset, with the time in seconds. Verify the expected forms of the heating+cooling and cooling alone sections. The heating term in T (t) is T heat (t) = 1 P (t) = 1 I R = I R t (10) where R is the resistance in the heating coil and the product I R can be taken outside the integral if the current I was constant during the heating and the resistance R does not depend significantly on the temperature of the coil in the temperature range considered. Also, no constant of integration has been written because this was already taken into account above in the expression for T (t) with both heating and cooling. Thus, the heating alone should yield a linear dependence on time; your plot should look roughly linear, but may show small deviations from linearity because what you are actually seeing during the first section of the data is the combined effect of the heating and cooling. We saw above that the cooling term yields an exponential temperature dependence: Θ cool (t c ) = Θ i e ktc T cool (t c ) = T o +Θ i e ktc (11) where t c is the time after the heating was switched off. You may not be able to see this dependence clearly on your graph if the exponential function is falling off slowly in time. Indeed, if k is small (the calorimeter is well insulated, so that the rate of cooling is slow), the exponential will be approximately equal to 1 kt c the first two terms in the Taylor expansion for e ktc.inthiscase, T cool (t c ) = T o +(T max T o )(1 kt c ) = T max +(T max T o )kt c (1) Thus, if the cooling is slow enough, the plot of temperature vs. time due to cooling should likewise be roughly linear. 4

5 The ooling oefficient k You have two distinct sections of data the first during the time you were heating the water in the calorimeter (with simultaneous cooling), and the second when the heating had been turned off and only cooling was occurring. You will first work with the latter part of the data, with cooling alone, in order to determine the cooling coefficient k by making a plot of ln Θ vs. t c (time since the heating was switched off) and finding the slope. 1. Make a new set of data columns containing only the data after the heating was switched off. Adjust the times so that the first time is zero and the other times are offset accordingly. Enter ln Θ in another column.. Plot ln Θ vs. t c and obtain a linear fit to these data to determine the cooling coefficient k. You should also derive an uncertainty for k. Your analysis package (e.g., Origin) may do this automatically. If not, you can determine the uncertainty in the slope k from the scatter of your data points about the linear fit.ifwehavefitalineoftheformy = mx + b, the uncertainty in the slope m is given by the formula (valid for precisely this situation uncertainties on each measurement roughly the same, reasonably large total number of data points): where σ m = N σ Δ (13) σ = 1 [yi (mx i + b)] (14) N Δ=N x i ( x i ) (15) N is the number of data points, and the sums go from i =1toN. Thequantity σ is essentially an estimate of the real uncertainty on the measured y values, based on the scatter of the points about the linear fit and taking into account the number of parameters we have fit (two). The uncertainty in the y-intercept is given by the formula: σb = σ x Δ i (16) where σ and Δ are the same as above. You can see from the equations in the introduction that the y-intercept of your line should be ln Θ i = T i T o ;your y-intercept should be equal to this value within the errors. 3. The cooling coefficient k = λa b λa = b(mc copper + Mc water ) (17) where λ =0.06 W m 1 K 1 is the thermal conductivity of dry air, A is the surface area of the cylindrical part of the calorimeter, b is the effective thickness 5

6 of the boundary layer of air and = mc copper + Mc water is the total heat capacity of the calorimeter, stirrer and water, where c copper = 385 J kg 1 K 1 and c water = 4180 J kg 1 K 1 are the specific heat capacities of copper and water. Solving for the thickness of the boundary layer: b = λa k(mc copper + Mc water ) (18) Use your value for k, your measured masses for the calorimeter + stirrer and water, the area of the calorimeter surface, and the other constants in this relation to estimate the thickness b of the boundary layer of air acting to conduct heat away from the calorimeter. Does your result seem reasonable? Finding a ooling orrection for the Data If we look again at the expression for the temperature as a function of time, we see that we can determine and subtract the cooling term in this expression, in order to reconstruct what T (t) =T meas would have looked like without the cooling occurring: T meas (t) = T o + 1 P (t) k (T (t) T o ) T no cool (t) = T meas + k (T (t) T o ) (19) where the integrals here are carried out from t =0tot = t. Although we do not have an expression for the integrand that we can use to directly evaluate the integral in the cooling term, we can evaluate this integral numerically by finding the area corresponding to (T (t) T o ), which will be the area on your plot of temperature vs. time below the measured temperature curve T (t) and above the curve for the background temperature as a function of time, which should be close to a horizontal line. Since you determined k above, you can find the overall cooling correction. 1. alculate the area under the temperature vs. time plot above the background temperature as a function of time by numerically integrating these data using the trapezoidal rule. If (x o,y o ), (x o + d, y 1 ), (x o +d, y ),...,(x o + nd, y n )are n + 1 equidistantly spaced data points (separation d), the total area under the curve joining the points is approximately given by ( y0 Area = d + y 1 + y + + y n 1 + y ) n (0) You can calculate the required area for the T (t) T o dependence as a function of time by first applying the above formula to the first time interval between measurements in this case, we have n + 1 = data points and n = 1 interval: Area(1) = d ( y0 + y ) 1 6

7 where d is the time interval between your measurements and y is T (t) T o.we will have for the area of the first two time intervals Area() = d ( y0 + y 1 + y ) = Area(1) + d (y 1 + y ) A little examination of the formula for the area will convince you that in general, Area(j) = Area(j 1) + d (y j 1 + y j ) (1) where Area(0) = 0 (start of the experiment). Use this relation to determine the area under your T (t) T o curve as a function of time.. The two sections on your plot of T (t) vs. time should both look approximately linear, for reasons discussed above. You could also estimate the total area under the curve by assuming the data in both sections are approximately linear relations, and simply calculating the areas for the two sections using the formula for the area of a triangle: half the base times the height. How well does this agree with the total area under the curve obtained using the trapezoidal rule? 3. You can now use the required area as a function of time obtained using the trapezoidal rule together with your value for k to correct the measured temperatures as a function of time for the Newtonian cooling occurring in both the heating+cooling and cooling alone sections of the data. 4. Make a plot showing both T meas (t) andt no cool (t), where the latter represents your measured temperatures after correcting them for the cooling. If this has been done correctly, your cooling-corrected plot should now show linear behaviour for the first section if your current was constant (heating alone) and a flat horizontal line for the second section (no heating or cooling). 5. Use the cooling-corrected curve to determine the maximum temperature that would have been reached during the heating if no cooling were occurring (i.e., if the calorimeter were perfectly isolated from the surroundings). alculate the heat energy delivered by the batteries to the water and calorimeter during the heating phase: Q = (c copper m + c water M)(T max T o ) () where c copper = 385 J kg 1 K 1 and c water = 4180 J kg 1 K 1 are the specific heat capacities of copper (the calorimeter and stirrer) and water, m is the mass of the calorimeter and stirrer and M is the mass of the water. 6. Now determine the emf V of the batteries from the relation Q = IV =< I>Vt heat (3) where < I > is the average current during the heating. Your result for V should be close to the emf deduced from the indicated voltages on the batteries. 7

EXPERIMENT ET: ENERGY TRANSFORMATION & SPECIFIC HEAT

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.01X Fall 2000 EXPERIMENT ET: ENERGY TRANSFORMATION & SPECIFIC HEAT We have introduced different types of energy which help us describe