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I am looking for an approximation here. I am assuming that the temperature change will be affected by the total surface area of water exposed to the warmer air.

I have a clear plastic container. It holds approximately 9 L of water. It is approximately cylindrical, maybe 30 cm in diameter

The water will be exposed to air at the top of the container.

I am also assuming that the starting temperature of the water will be approximately 2-5 C, and will finish at 22 C.

Can the specific heat of water be used to calculate the time needed?
What is the time needed to warm the water?
What is the physics and mathematics behind the solution?

I've studied both calculus and physics, but it has been so long that I can't remember how to do a calculation like this.

2007-03-01 23:21:45 · 2 answers · asked by Beam 3 in Science & Mathematics Physics

2 answers

Cannot really solve this problem with only those givens, because the heat transfer curve cannot be determined.

The water temperature will asymptotically approach that of the room following an exponential curve. You only need to find the time constant of this exponential curve. During the elapse of one time constant, the system crosses (1-1/e) = 63% of the remaining temperature gap.

In this case if the initial temp is 2C and the final is 20C, after the first elapse of one time constant the temperature will be (20 - 2) * .63 + 2 = 13.3C. If you measure the time from the beginning until the water temperature is 13.3C average, that will be your system's time constant. And from there on, after every time constant the remaining temperature gap will shrink by 63% until the asymptotic gap is within your tolerance.

2007-03-01 23:54:38 · answer #1 · answered by sciquest 4 · 0 0

ive never studied calculas or physics but i'd imagine that you would need to know the room temperature, the water temperature from the tap, current humidty and maybe even braometric pressure which one would think all would play a key role in this experiment. ie if its a hot day it may take 5 mins but a cooler day may take 15 depending on the temp of the water when it leaves the tap. although you mention that the air will hit the water at the top dont forget to factor into ur calculations the type of plastic ur container is made from. the density of the plastic will alter the time it takes to warm the water up as some thinner plastics may not be as great an insulator as a thicker plastic. so you now have an extra variable involved in ur experiment to take into consideration. you would also need to know outside temp for same reason as with the plastic. if the day is forecast to stay fairly even then it should be a steady rise in temp but if the temp varies even by 5 degrees it could change the room temp and affect the speed of the water being heated. (depends on insulation of the room)

water temp from tap
room temp
humidity
barometric pressure
outside temp
plastic quality

i hope this makes sense.

2007-03-01 23:50:15 · answer #2 · answered by jamie28981 2 · 0 0

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