Can anyone help me with this problem?
Suppose that during a controlled experiment, the temperature in a beaker containing some chemical substance at time t is rising at a rate of (15/13)t^2-(7t) degrees centigrade per minute. If the temperature after 1 minute is measured to be 18.6 degrees C, what is the temperature in the test tube after 10 minutes?
If you integrate the rate, you get temp.
Temp= INT (15/13 t^2 -7t)dt
You know temp at 1 min is 18.6C. Solve for the constant of integration.
Then, you can find the temp after ten min.
To solve this problem, we can use integration to find the equation for the temperature in terms of time, and then substitute t = 10 minutes into the equation to find the temperature after 10 minutes.
The given rate of change of temperature as a function of time is:
dT/dt = (15/13)t^2 - 7t
To find the equation for the temperature T in terms of time t, we can integrate both sides of the equation:
∫ dT = ∫ [(15/13)t^2 - 7t] dt
Integrating the right side of the equation:
T = ∫ (15/13)t^2 dt - ∫ 7t dt
Simplifying the integrals:
T = (15/39)t^3 - (7/2)t^2 + C
Where C is the constant of integration.
To find the specific value of C, we can use the information given in the problem. At time t = 1 minute, the temperature T is measured to be 18.6 degrees Celsius. Substituting these values into the equation:
18.6 = (15/39)(1)^3 - (7/2)(1)^2 + C
Simplifying the equation:
18.6 = 15/39 - 7/2 + C
To find the value of C, combine the fractions and solve for C:
C = 18.6 - 15/39 + 7/2
C = (5514 - 585 + 11313)/78
C = 16542/78
C = 211.6154
Now we have the equation for the temperature T in terms of time t:
T = (15/39)t^3 - (7/2)t^2 + 211.6154
To find the temperature after 10 minutes, substitute t = 10 into the equation:
T = (15/39)(10)^3 - (7/2)(10)^2 + 211.6154
Simplifying the equation:
T = (1500/39) - (350/2) + 211.6154
T = 38.4615 - 175 + 211.6154
T = 75.0769 degrees Celsius
Therefore, the temperature in the test tube after 10 minutes is approximately 75.08 degrees Celsius.