An insulated tank is filled with a solution containing radioactive cobalt. Due to the radioactivity, energy is released and the temperature T (in deg C) of the solution rises with time t (in h). The following equation expresses the relation between temperature and time for a specific case: dT=(20.0-2.50t)dt. If the initial temperature is 65 degrees C, what is the temperature 12 hours later?
2007-09-05 06:37:19 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
T= 20t -1.25t^2 +C
When t=0, T= 65, so C = 65
T= 20t - 1.25t^2 +65
T= 20(12) - 1.25*12^2 +65
T = 240 - 180 +65 = 125 degrees C
2007-09-05 06:48:44 · answer #1 · answered by ironduke8159 7 · 0⤊ 1⤋
I don't get it, and maybe you're having the same problem. T is the temperature of the solution and it varies with time and we have dT = (20 - 2.5t) dt. So T at any time is given by the integral of 20 - 2.5t between limits to be decided upon. In this case we want between 0 and 12. It seems to me the answer should be 20t - 1.25 t^2 evaluated at 12 minus evaluated at 0.
Or 240 - 180 = 60. This doesn't make use of the initial temp. And supposing the initial temp. was 65, the temp after 12 hours has dropped, making no sense. Are you sure T represents the temperature and not the increase in temperature?
2007-09-05 14:15:10 · answer #2 · answered by rrsvvc 4 · 0⤊ 1⤋
dT = (20,0 - 2,50t)dt (integral)
T = 20,0t - 1,25t^2
the temperature for 12 hours
T = 20,0 . 12 - 1,25 . (12)^2
T = 240 - 180
T = 60 deg C
because the initial temperature is 65 deg C so the temperature 12 hours later is
65 deg C + 60 deg C
125 deg C
2007-09-05 14:26:35 · answer #3 · answered by fortman 3 · 0⤊ 0⤋