Help MathCad 15: water tank w/ leakage rate prop. to sqrt of water column

[Saila]

Hello, Please help to solving two tasks with MathCad?

1) Water tank, the base of which runs through the water, is filled with 50 cm heights. It is known that the leakage rate is proportional to the square root of the water column. During hour, the water level dropped 9 cm. How long is the water level dropped to 40 cm and when the tank is empty?

2)During the test, the stone was placed at a temperature of 60 ° in the current form at 20 degrees in water. The stone was cooled to 40 ° C for 2 minutes. Find the temperature of the stone for 5 and 10 minutes after.

Graphs is needed

Please feedback

Br.V

 

[HallsofIvy]

You don't say what the shape of the water tank is. I am going to assume a vertical circular cylinder. The "leakage" is the rate of change of the volume with respect to time and, with constant cross section area, is proportional to the rate of change of the height, dh/dt. Saying the rate of change of the volume is proportional to the square root of the height means that \(\displaystyle \frac{dh}{dt}= Ch^{1/2}\) where C is the "constant of proportionality".
cC
That "differential equation" can be separated as \(\displaystyle h^{-1/2}dh= Cdt\). Integrate both sides of that. Of course that will have a "constant of integration". Determine the two constants of from the condition that, initially the height was 50, h(0)= 50, and after one hour the water level has dropped 9 cm., h(1)= 50- 9= 41.

For the second problem, use "Newton's law of cooling". A warm object placed in a cool environment cools at a rate proportional to the difference in temperatures. Let the temperature of the stone, at time t, be T(t). Then dT/dt= C(T- 20) where, again, C is the "constant of proportionality". Write that as dT/(T- 20)= C dt and integrate both sides. Again you will have two unknown constants, C and the constant of integration. Use the facts that T(0)= 60 and T(2)= 40 to determine those constants.