Newton’s Law of Cooling

Objectives Use Newton’s Law of Cooling to determine the time of death in a controlled environment.

Temperature Change Newton’s Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the temperature of its surroundings (i.e. the ambient temperature ).

Temperature Change An object’s temperature over time will approach the temperature of its surroundings The greater the difference between the object’s temperature and the temperature of its surroundings, the greater the rate of change of the object’s temperature This change is a form of exponential decay T0 Ts

Newton’s Law of Cooling The rate at which an object cools is proportional to the difference in temperature between the object and the temperature of its surrounding: Where T is the final temperature of the object TO is initial temperature of the object k is a cooling constant TS is the temperature of the surroundings t is time A coroner uses this to help determine the time of death and is seen in every “Crime” TV series from Dragnet to CSI. T – TS = (TO – TS) e-kt Teacher Make sure that they notice that T is now Tlater and TO is now Tdisc and TS is now Tamb and t is now Δt and that k has not changed

Newton’s Cooling Equation Given by solving the equation for -kt we can begin to find the cooling constant for the system (the body and its surroundings) The symbol “𝑙𝑛” refers to the “natural log”. The only thing you have to know about it is where the button is on your calculator 𝑘𝑡=− 𝑙𝑛 𝑇− 𝑇 𝑆 𝑇 𝑂 − 𝑇 𝑆 If both sides are multiplied by 1/t the cooling constant can be calculated. 𝑘=− 1 𝑡 𝑙𝑛 𝑇− 𝑇 𝑆 𝑇 𝑂 − 𝑇 𝑆 T – TS = (TO – TS) e-kt IF you have a class set of calculators the teacher should find where the “𝑙𝑛“ button is on them and figure out if the number has to be entered before the“𝑙𝑛“button of after. If the students are suing their own some time should be given to let them explore their calculators to find out to find the button and how to use it.

Example 𝑘=− 1 𝑡 𝑙𝑛 𝑇− 𝑇 𝑆 𝑇 𝑂 − 𝑇 𝑆 𝑘=− 1 45 𝑙𝑛 300−72 350−72 Example: A potato is taken out of a 350o F oven and left to cool in a room at 72o F. The potato has cooled to a temperature of 300o F in 45 minutes. Find the value of k. Use the equation solved for k Plug in your values for each variable and perform each operation on your calculator. The value for k (cooling constant) for this potato in these surroundings is: 𝑘=− 1 𝑡 𝑙𝑛 𝑇− 𝑇 𝑆 𝑇 𝑂 − 𝑇 𝑆 𝑘=− 1 45 𝑙𝑛 300−72 350−72 𝑘=0.0044061219 7

Forensics Use of Newton’s Equation Now the variables have to be changes to match the field of forensics. Now: Tlater is the temperature of the body at the second reading Tdisc is initial temperature when the body was discovered k is a cooling constant Tamb is the ambient temperature of the room (surroundings) Δt is the time difference between the first and second temperature readings What do you notice about the color coding from this set of variables and the variables used earlier in the presentation.

Forensics Use of Newton’s Equation The formula used in forensics now looks like this: Where: Tlater is the temperature of the body at the second reading Tdisc is initial temperature when the body was discovered k is a cooling constant Tamb is the ambient temperature of the room (surroundings) Δt is the time difference between the first and second temperature readings 𝑘=− 1 ∆𝑡 𝑙𝑛 𝑇 𝑙𝑎𝑡𝑒𝑟 − 𝑇 𝑎𝑚𝑏 𝑇 𝑑𝑖𝑠𝑐 − 𝑇 𝑎𝑚𝑏

Forensics Use of Newton’s Equation Now that k can be solved, we use that value and plug it back in to solve for time of death. Take the formula Solve for ∆𝑡 Substitute in the variable (Δtdeath) for (Δt) and substitute he variable (Tdeath) for (Tlater). So…… 𝑘∆𝑡=− 𝑙𝑛 𝑇 𝑙𝑎𝑡𝑒𝑟 − 𝑇 𝑎𝑚𝑏 𝑇 𝑑𝑖𝑠𝑐 − 𝑇 𝑎𝑚𝑏 ∆𝑡=− 1 𝑘 𝑙𝑛 𝑇 𝑙𝑎𝑡𝑒𝑟 − 𝑇 𝑎𝑚𝑏 𝑇 𝑑𝑖𝑠𝑐 − 𝑇 𝑎𝑚𝑏

Forensics Use of Newton’s Equation Now the equation is set up to find change in time since death substitute in the variable (Δtdeath) for (Δt) and substitute the body temperature at the time of death 98.6o F (Tdeath) for (Tlater). So…… Becomes: Once the value of k is found simply plug in the data collected and solve with your calculator. ∆𝑡=− 1 𝑘 𝑙𝑛 𝑇 𝑙𝑎𝑡𝑒𝑟 − 𝑇 𝑎𝑚𝑏 𝑇 𝑑𝑖𝑠𝑐 − 𝑇 𝑎𝑚𝑏 ∆ 𝑡 𝑑𝑒𝑎𝑡ℎ =− 1 𝑘 𝑙𝑛 𝑇 𝑑𝑒𝑎𝑡ℎ − 𝑇 𝑎𝑚𝑏 𝑇 𝑑𝑖𝑠𝑐 − 𝑇 𝑎𝑚𝑏

Forensics Use of Newton’s Equation Assume that the value of k is .132647814 (do not round this number) and the body temperature at the time of discovery was 90.4o F with the temperature of 680 Find the time since death. Plug-in: Results in: This number will let you know how many minutes or hours ago the person died. ∆𝑡=− 1 𝑘 𝑙𝑛 𝑇 𝑑𝑒𝑎𝑡ℎ − 𝑇 𝑎𝑚𝑏 𝑇 𝑑𝑖𝑠𝑐 − 𝑇 𝑎𝑚𝑏 ∆ 𝑡 𝑑𝑒𝑎𝑡ℎ =− 1 0.132647814 𝑙𝑛 98.6−68 90.4−68 Pose the question….. Which units do you think should be used here? (Will a body really cool down that much in two minutes in a 68 degree room.) You are trying to get them to understand that they can’t just blindly accept the answer. I must align with the “real world science”. ∆ 𝑡 𝑑𝑒𝑎𝑡ℎ =−2.352

Newton’s Law of Cooling Example: The great detective Sherlock Holmes and his assistant, Dr. Watson, are discussing the murder of actor Cornelius McHam. McHam was shot in the head, and his understudy, Barry Moore, was found standing over the body with the murder weapon in hand. Let’s listen in: Watson: Open-and-shut case, Holmes. Moore is the murderer. Holmes: Not so fast, Watson – you are forgetting Newton’s Law of Cooling! Watson: Huh? Holmes: Elementary, my dear Watson. Moore was found standing over McHam at 10:06 p.m., at which time the coroner recorded a body temperature of 77.9°F and noted that the room thermostat was set to 72°F. At 11:06 p.m. the coroner took another reading and recorded a body temperature of 75.6°F. Since McHam’s normal temperature was 98.6°F, and since Moore was on stage between 6:00 p.m. and 8:00 p.m., Moore is obviously innocent. Ask any calculus student to figure it out for you. How did Holmes know that Moore was innocent? After reading through this slide, have them copy down the data on a new sheet of paper and solve for k and ∆ 𝒕 𝒅𝒆𝒂𝒕𝒉 Use this as an exit ticket as a formative assessment.