Ive been trying to work this question out on my own for over a week now and cant seem to crack it. I hope its in the right Sub forum and im really sorry if its not.
A hot metal body is allowed to cool in still air and the difference in temperature, θ (Celsius) between the body and its surroundings is recorded at, t (seconds) from the beginning of the cooling period.
The question asks me to verify graphically the relationship between θ and t is of the form: θ=θoe-kt (-kt is to the power of θoe) Hence estimate the values of the constants θo and k. Then using this relationship compile a table of values of θ for values of t between 0 and 600 at intervals of 60.
I have taken the natural log (ln) of each of the temperatures in the table above and plotted them in the graph. I drew a line of best fit and found the y intercept, and took the natural exponential of it which is:
e4.25 = 70.11
I think 70.11 is the initial temperature at time 0. so θo= 70.11?
Is k the gradient of the line? if so how do i calculate the values of θ?
All help is greatly appreciated, it would be amazing to get this out of the way!
March 10th 2011, 05:04 AM
Newton's law says that . Take natural logarithms of that equation: . So what you should do is to graph the logarithm of theta against t. The graph should then look like a straight line, with gradient ľk and intercept . I haven't tried to check the numbers, but it looks as though that is what you have done. If so, you're right!
Once you have the values for and k, you can plug them into the original formula , and use it to find for any given value of t.
March 10th 2011, 05:27 AM
Ok i think im right so far. All i need now is to estimate the values of the constants θo and k.
I have worked out
θo = 70.11 and k = -0.004
Now to compile a table of values of θ for values of t between 0 and 600 at intervals of 60.
If and , then your formula is . For example, when t = 120, is given by
But there's something wrong with this! It says that after 120 seconds, the temperature difference between the hot object and its surroundings (43.4 degrees) is greater than it was at t=60 seconds, namely 40.5 degrees according to the given data. In other words, the object has started to heat up again!
Once you see that this can't be correct, you should look back to check the values that you got for and k. The value of is the temperature difference at time 0, which should obviously be greater than its value at any later time. But you are told that when t=10. So the value is very wrong. I should think it would be about ten times that.
March 10th 2011, 07:23 AM
Here's what I got - I think it looks pretty reasonable.