Lab 3- Newton's Law of Cooling

Introduction: Newton’s law of cooling was experimentally observed after heating a copper tube and recording the drop in temperature with time. The recorded times and natural log of the changes in temperature were used to make a graph, which allowed for the determination of the time constant 𝜏. Theory: According to Newton’s law of cooling, the rate of heat flow of an object is proportional to the temperature difference between the temperature and its surroundings. dQ/dt = -k∆T From this equation K is a constant and ∆T = T-Ts, in which Ts is the temperature of the surrounding air. After differentiating with respect to time from the equation Q=mC∆T, the following equation arises: dQ/dt=mc(dT/dt) When the right hand side of this equation is applied to the first equation of dQ/dt=-k∆T, then the following equation arises: mc(dT/dt)=-k∆T, which can be manipulated into dT/∆T=(-k/mc)dt. Both sides of the equation are then integrated: 𝑇

𝑡

𝑘

∫𝑇𝑜 𝑑𝑇/(𝑇 − 𝑇𝑠)=∫0 (− 𝑚𝑐 ) 𝑑𝑡  Ln(T-Ts)- Ln(To-Ts)= -kt/mc  Ln[(T-Ts)/(To-Ts)]=-kt/mc Take both sides of the equation to the power of e: (T-Ts)/(To-Ts)=e-kt/mc Then Multiply both sides of the equation by (To-Ts): T-Ts=(To-Ts)e-kt/mc

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∆T=(To-Ts)e-kt/mc

Because 𝜏 = mc/k, the equation simplifies to ∆T=(To-Ts)e^(-t/ 𝜏)

Procedure: Assemble the station in accordance with the diagram of the lab manual. Ensure that the thermometer is not touching the sides of the copper tube. Rather, it should be directly centered inside of the tube. Heat the tube with a propane torch uniformly along its length to a temperature between 125ºC and 150ºC. After the temperature has stabilized, it shall be recorded as To. The temperature shall be recorded at 15 second intervals for the first two minutes, then at 30 second intervals for the next seven minutes.

Analysis and Discussion: After viewing the graph of Ln∆T vs. time, it is evident that the trendline has a greater slope towards the far left of the graph, which consists of higher temperatures and lower time values. On the other hand, there is less slope towards the far right of the graph, which consists of lower temperatures and higher time values. The slope represents the rate of change of heat loss. According to Newton’s law of cooling, the rate of heat loss of a cooling body is proportional to the difference in temperatures between the body and its surroundings. The far left of the graph has a greater difference between the tube’s temperatures and the surrounding temperature, while the far right of the graph contains smaller differences between the tube’s temperature and the surrounding temperature. Because the far left of the graph has greater slope than the far right of the graph, it may be concluded that the rate of heat loss is proportional to the difference between the tube and the surrounding temperature. Thus, the experiment confirmed Newton’s law of cooling. The maximum slope occurs between 150 and 180 seconds, producing a slope of -0.00191. The minimum slope appears to appear at the very bottom right of the graph, between 840 and 810 seconds, which produces a slope of -0.0010. From Microsoft Excel’s linear trendline, the best fit slope resulted in -0.0027. These slope values were used to find the % slope error: % slope error= (max slope-min slope)/2(best fit slope) = (-.00191+.0010)/[2(-.0027)] = 16.85% The slope error of 16.85% suggests that either personal error or experimental equipment may have provided faulty results. From the equation ∆T=(To-Ts)e^(-t/ 𝜏), taking the natural log on both sides produces: Ln∆T=Ln(To-Ts)-t/ 𝜏 This equation reflects the slope-intercept form of the equation y=mx+b. Because time is the xcoordinate on the graph, -1/ 𝜏 is the slope of the graph. 𝜏 may be calculated by equating -1/ 𝜏 to the slope from the graph. -1/ 𝜏=-0.0027

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𝜏=1/0.0027=370.4

Conclusion: The experiment provided for successful observation of Newton’s law of cooling. From the graph, it is evident that the temperature of the body declines as time increases. The graph also illustrates that the rate of the tube’s heat loss is proportional to the temperature of the tube and the surrounding temperature. The slope of the graph allowed for the determination of 𝜏 = 370.4. The 16.85% slope error may be attributed to slight faults in the equipment used in the experiment. It is possible that the thermometer used may have not recorded the exact temperature of the copper tube, for the thermometer was not in direct contact with the tube but was rather measuring the air temperature just near the copper. This may have provided slightly incorrect measurements of the tube that may have disrupted the accuracy of finding slope from the experiment’s equations. A thermometer that can record temperature by direct contact with the thermometer may possibly provide more accurate measurements. Also, personal error may have attributed to the 16.85% slope error. Exact temperatures may have not been recorded at the exact time intervals of 15 and 30 seconds, which may have provided just enough of measurement errors to disrupt the accuracy in calculating slope.

Newton’s Law of Cooling

The experiment was focused to observe Newton’s law of cooling and to determine the time constant 𝜏 from the equation ∆T=(To-Ts)e^(-t/ 𝜏). Tools used in the experiment included a propane torch, a copper tube, a thermometer, and a stopwatch, all of which were used to record temperature decrease of the copper tube in relation to time. The measurements provided for a graph of Ln∆T vs. t, where ∆T is the temperature difference between the copper tube and the surrounding air temperature and t is time. The graph provided a slope that could be used to find 𝜏=370.4 The graph was also used to find a slope error of 16.85%. The varying slopes of the graph verified Newton’s law of cooling, in which the rate of heat loss of a cooling body is proportional to the difference in temperatures between the body and its surroundings.

William Drury Alejandro Bolanos

PHYS M20BL-32487 4 February, 2015

Questions: 1. Ln[(T-Ts)/(To-Ts)]=-t/ 𝜏 To-Ts=122.5; 1 % of (To-Ts)=121.275  t=- 𝜏 Ln[(T-Ts)/(To-Ts)]= 370.4*Ln(121.275/122.5)= 3.72 seconds 3. Because the initial temperature To is in the equation t=- 𝜏 Ln[(T-Ts)/(To-Ts)], time must depend on intial temperature.