It's there because of the exponential decay, but if you don't put it there, then the constant will just work out to be negative for cooling anyway. Notice in my first example I used T - Ts = (To - Ts)e^kt and k = -0.06108. If I had used T - Ts = (To - Ts)e^-kt, then k would be 0.06108.
@11031992em The exponent is -kt since this is exponential decay (not growth). Some people leave out the negative and then the k value will end up being negative anyway, so it is doesn't really matter whether it's in the formula or your k number is < 0.
Here's my solution: T - 0 = (115 - 0)e^kt so basically T = 115e^kt At 5 min: 80 = 115e^5k dividing by 115 we get e^5k = 80/115 or e^5k = 16/23 Solving for k we get 5k = ln(16/23) and then dividing by 5 to get k = (1/5)ln(16/23) = -0.0726 So T = 115e^-.0.0726t at 10 min T = 115e^-0.0726(10) = 55.7 degrees F.
So when we're trying to find out how long cooling will take, we'll need that -k in the denominator of a natural log's argument. Problem is, logs can't have a negative argument. I'm having that problem right now in Algebra and I'm confused. I'm assuming we need the numerator of the argument to be negative so the two negative signs cancel, but I can't figure out how. Any ideas?
There's no calculus necessary in using Newton's Law of Cooling. It just states a relationship between ambient temperature, time and the temperature at a certain time.
I have this equation that i cant seem to solve. A latte with a temperature of 115 degrees F is placed in a freezer of 0 degrees F. After five minutes the latte was at 80 degrees F. What will the temp be in 10 minutes. I tried following the instructions but I must be doing something wrong Please and Thank You anyone!
The negative in Newton's Law of Cooling is a traditional negative, and is by far one of the most irritating things. Why would they put it there? Who knows, just to cause commotion. Leave it in there to not cause confusion upon doing heating up problems. Just know that the sign will be reversed in correlation with it heating/cooling.
When solving for t you should not end up having/trying to take the log of a negative number. In my first example from the video it had 5 = 95e^-0.6108t then dividing both sides by 95 we get 1/19 = e^-0.6108t so to convert to log form it becomes -0.6108t = ln(1/19) so t = ln(1/19)/-0.6108 to get the time. If this isn't making sense then email me your question so I can respond specifically to it. My email address is al.richards@kpdsb.on.ca.
It makes no difference what so ever whenever the temp of the WATER is at 176 but no higher than 181 Fahrenheit but these are water estimates it's just a 32 ounce clear water pitcher unfilled my point a major issue of water temp occurred cold water 75 farenhoent left in room temp in the same type of pitcher at the same amount of time 90 minutes a 32 ounce pitcher of water not frozen or boiling will be close to the rooms temperature especially if it's there for 90 minutes equally 32 fluid ounces of water and with 20000 square ft not including the breeze
this can be done by an easier way which would reduce this sum to a a matter of addition and subtraction. no need to take out ln and other difficult division.
It's there because of the exponential decay, but if you don't put it there, then the constant will just work out to be negative for cooling anyway. Notice in my first example I used T - Ts = (To - Ts)e^kt and k = -0.06108. If I had used T - Ts = (To - Ts)e^-kt, then k would be 0.06108.
this helped me out so much thank you my good man as i have a quiz in 30 min
This clip helped me out a bunch. Thanks a lot!
Thank goodness for you, you saved the day!
@11031992em The exponent is -kt since this is exponential decay (not growth). Some people leave out the negative and then the k value will end up being negative anyway, so it is doesn't really matter whether it's in the formula or your k number is < 0.
Here's my solution:
T - 0 = (115 - 0)e^kt so basically T = 115e^kt
At 5 min: 80 = 115e^5k
dividing by 115 we get e^5k = 80/115 or e^5k = 16/23
Solving for k we get 5k = ln(16/23) and then dividing by 5 to get k = (1/5)ln(16/23) = -0.0726
So T = 115e^-.0.0726t at 10 min T = 115e^-0.0726(10) = 55.7 degrees F.
sir, i have a formula here sayin
T= Tm + (To - Tm) e^(-kt)
why is k negative? x_X
So when we're trying to find out how long cooling will take, we'll need that -k in the denominator of a natural log's argument. Problem is, logs can't have a negative argument. I'm having that problem right now in Algebra and I'm confused. I'm assuming we need the numerator of the argument to be negative so the two negative signs cancel, but I can't figure out how. Any ideas?
I have a doubt that why cant we use the equation of dT/dt = k(T-To) directly for the rate. By integrating this only we got the above used equation.
There's no calculus necessary in using Newton's Law of Cooling. It just states a relationship between ambient temperature, time and the temperature at a certain time.
Figured it out. Thanks so much!
Awesome, very helpful.
Nice video I'm learning something similar except it's specific heat and latent heat equations.
I have this equation that i cant seem to solve.
A latte with a temperature of 115 degrees F is placed in a freezer of 0 degrees F. After five minutes the latte was at 80 degrees F. What will the temp be in 10 minutes. I tried following the instructions but I must be doing something wrong
Please and Thank You anyone!
What program did you use to create this video?
Hi,
The examples were created in a PowerPoint and then recorded using a screen capture application called Camtasia 7, made by Techsmith.
Al Richards thank you!!
thank you
Thanks
The negative in Newton's Law of Cooling is a traditional negative, and is by far one of the most irritating things. Why would they put it there? Who knows, just to cause commotion. Leave it in there to not cause confusion upon doing heating up problems. Just know that the sign will be reversed in correlation with it heating/cooling.
When solving for t you should not end up having/trying to take the log of a negative number. In my first example from the video it had 5 = 95e^-0.6108t then dividing both sides by 95 we get 1/19 = e^-0.6108t so to convert to log form it becomes -0.6108t = ln(1/19) so t = ln(1/19)/-0.6108 to get the time.
If this isn't making sense then email me your question so I can respond specifically to it. My email address is al.richards@kpdsb.on.ca.
It makes no difference what so ever whenever the temp of the WATER is at 176 but no higher than 181 Fahrenheit but these are water estimates it's just a 32 ounce clear water pitcher unfilled my point a major issue of water temp occurred cold water 75 farenhoent left in room temp in the same type of pitcher at the same amount of time 90 minutes a 32 ounce pitcher of water not frozen or boiling will be close to the rooms temperature especially if it's there for 90 minutes equally 32 fluid ounces of water and with 20000 square ft not including the breeze
this can be done by an easier way which would reduce this sum to a a matter of addition and subtraction. no need to take out ln and other difficult division.
its take 28.6 min not 48.6 min .
I believe the 48.6 min is correct. How did you get 28.6?
Vengo por Mar 🤪
this dont work sir
What's not working?
AlRichards314 doesn’t work with my question for some reason :(
@@frostbitepokin9520 Hey. Tell me what your question is so I can take a look at it.
AlRichards314 can I send you a picture?
@@frostbitepokin9520 Yes, sure email it to alrichards314@gmail.com