Question

A dead body is found on a cold, 15°C, winter night at 1 am. The coroner knows that due to the cold

biological process slow down significantly. At 30°C the rate at which a body decomposes is .05% per hour
(see graph). If the body is at .07% what time did the person die, assuming it as been 15°C since 7pm.

Answer 1

Answer with Explanation:

Let's call the time that the person died "t" in hours since 7pm. From 7pm to 1am is 6 hours, so t+6 is the total number of hours since the start of the cooling process.

At 30°C, the rate of decomposition is 0.05% per hour, or 0.0005. At 15°C, we know that the rate of decomposition is slower, but we don't know the exact value. Let's call it "x" for now.

Using the information given, we can set up an equation to solve for t:

0.07 = (t+6)(0.0005)(x)

We can simplify this to:

x = 0.07 / (0.0005)(t+6)

We also know that the time since the start of the cooling process (t+6) plus the time between the cooling process and when the body was found (5 hours) is equal to the total time since the person died:

t+6+5 = 8

Simplifying this, we get:

t = -3

This means that the person died 3 hours before the start of the cooling process, which was at 7pm. Therefore, the person died at 4pm.

Hope it helps! : )