Time of Death: Newton's Law Versus a Linear Cooling Shortcut
A police team finds a body in a room whose air temperature is steady at 20°C. At 10:00 pm the body temperature is 30°C, and at 11:00 pm it is 28°C. The investigator assumes Newton's law of cooling, dT/dt = -k(T - 20), where T is the body temperature in °C and t is in hours after 10:00 pm. The normal living body temperature is taken as 37°C. A junior officer argues that because the body lost 2°C in the first hour, it would have lost 2°C per hour all along, so death occurred 3.5 hours before 10:00 pm. The investigator is unsure whether the cooling rate is constant in that simple way. The team also wants to know when the body temperature would reach 25°C, and whether the estimate is sensitive to the assumption that the room stayed at 20°C. You are asked to solve the model, estimate the time of death, and evaluate the junior officer's linear shortcut.
1. Solve the differential equation with T(0) = 30 and T(1) = 28 to find k, and write T(t).
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Separate variables or use the standard solution:
dT/(T - 20) = -k dt, so T - 20 = (T(0) - 20)e^(-kt).
With T(0) = 30, T(t) = 20 + 10e^(-kt). At t = 1, T = 28:
28 = 20 + 10e^(-k), so e^(-k) = 0.8.
Thus k = -ln 0.8 ≈ 0.2231 h⁻¹, and
T(t) = 20 + 10e^(-0.2231t) = 20 + 10(0.8)^t,
where t is in hours after 10:00 pm.
2. Estimate the time of death using T = 37°C, and find when T = 25°C. Give the times relative to 10:00 pm.
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For T = 37:
37 = 20 + 10e^(-kt), so e^(-kt) = 1.7.
Then -kt = ln 1.7, giving t = -ln 1.7/k = -0.5306/0.2231 ≈ -2.38 h. The negative sign means death occurred before 10:00 pm. Since 0.38 h ≈ 23 min, death was around 7:37 pm.
For T = 25:
25 = 20 + 10e^(-kt), so e^(-kt) = 0.5.
Then t = ln 2/k = 0.6931/0.2231 ≈ 3.11 h after 10:00 pm, which is about 1:06 am. Check: T(3.11) = 20 + 10(0.8)^3.11 ≈ 20 + 10(0.5) = 25°C.
3. Critically evaluate the junior officer's 3.5-hour estimate. State the direction of the error and one practical reason the model may mislead in a real room.
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The junior officer assumes a constant cooling rate of 2°C per hour. Newton's law gives a rate proportional to T - 20, so the body cools faster when it is hotter and more slowly as it approaches room temperature. To lose the first 7°C (37°C to 30°C), the excess temperature is larger than during the 30°C to 28°C interval, so that stage takes less than 3.5 h. The exponential model estimates about 2.38 h, so the linear shortcut overestimates the post-mortem interval.
In a real room, the estimate can mislead because the room may not stay at 20°C, the body is not a uniform lump, clothing and contact with the floor alter heat loss, and k estimated from just one hour is uncertain. The result should be reported as approximate and checked against other evidence.