Finding time of death using Newton's cooling law

[Psyguy22]

1 the problem and all known variables.
A person is murdered in a room with a temperature of 20 deg C. At the time the body is discovered, the body temp is 32 deg C and is decreasing at an instantaneous rate of .1 deg C/minute. How long ago was the murder commited?

2. Related equations.
Normal body temp is 37 deg C
Y(t)=T+Ae^(-kt) where y(t) is temp at a given time, T is the room temp, and A and K are constants related to cooling.

3 attempt at solution.
So my teacher gae the hint to have t=0 be the time the body was discovered. So you'd get 32 (y(0)) = 20+A ((e^0=1). I got that A=12. So plug that back into the eq. And because the body temp decreases at .1 deg C/minute, K would equal .1. So then I solve y(t) for 37 and I get -3.7 which if it were in hours, that answer would make sense. But since k was in minutes, I figured my answer was too so then I converted k to hours revised my answer and got -207 which is a really long time. I'm confused in where I went wrong.

 

[Simon Bridge]

Did you make sure your units were consistent?

Notes:
An instantaneous decrease rate of 0.1deg/min at some time means dY/dt at that time was -0.1/min
You do not have a variable K in your equation - important to be consistent in your notation.

 

[Psyguy22]

So I just solve dy/dt= -.1 when t=0 and that should give me my k?

 

[Simon Bridge]

That's how I read it.

 

[Ray Vickson]

1 the problem and all known variables.
A person is murdered in a room with a temperature of 20 deg C. At the time the body is discovered, the body temp is 32 deg C and is decreasing at an instantaneous rate of .1 deg C/minute. How long ago was the murder commited?

2. Related equations.
Normal body temp is 37 deg C
Y(t)=T+Ae^(-kt) where y(t) is temp at a given time, T is the room temp, and A and K are constants related to cooling.

3 attempt at solution.
So my teacher gae the hint to have t=0 be the time the body was discovered. So you'd get 32 (y(0)) = 20+A ((e^0=1). I got that A=12. So plug that back into the eq. And because the body temp decreases at .1 deg C/minute, K would equal .1. So then I solve y(t) for 37 and I get -3.7 which if it were in hours, that answer would make sense. But since k was in minutes, I figured my answer was too so then I converted k to hours revised my answer and got -207 which is a really long time. I'm confused in where I went wrong.

The sentence " At the time the body is discovered, the body temp is 32 deg C and is decreasing at an instantaneous rate of .1 deg C/minute" seems to be saying that the 0.1 deg C/min applies at the time the body is discovered, if we read it as a standard English sentence. This gives two equations in the two unknowns k and t. Solving it gives a much more "reasonable" value for t (keeping t in minutes throughout). Try it and see!

 

[Ray Vickson]

1 the problem and all known variables.
A person is murdered in a room with a temperature of 20 deg C. At the time the body is discovered, the body temp is 32 deg C and is decreasing at an instantaneous rate of .1 deg C/minute. How long ago was the murder commited?

2. Related equations.
Normal body temp is 37 deg C
Y(t)=T+Ae^(-kt) where y(t) is temp at a given time, T is the room temp, and A and K are constants related to cooling.

3 attempt at solution.
So my teacher gae the hint to have t=0 be the time the body was discovered. So you'd get 32 (y(0)) = 20+A ((e^0=1). I got that A=12. So plug that back into the eq. And because the body temp decreases at .1 deg C/minute, K would equal .1. So then I solve y(t) for 37 and I get -3.7 which if it were in hours, that answer would make sense. But since k was in minutes, I figured my answer was too so then I converted k to hours revised my answer and got -207 which is a really long time. I'm confused in where I went wrong.

Please ignore my previous response; it is based on a mis-reading of what you did. (I tried to edit or delete it, but I guess too much time has passed and so those options are now void.) The problem is your incorrect value of k: you need $y'(0) = -1/10$ (according to the problem's statement), so $12 k = 1/10$.