Asked by Alice
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?
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?
Answers
Answered by
bobpursley
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.
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.
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