Math help concerning heat and joules or Specific Heat Capacity equation?
Reach
Posts: 107
Hello,
I will admit word problems are my worst when dabbling in math so I need a little help.
Lets assume I have a 5000 mm volume of water resting at 25 c. I want to rise the temperature to 100 c in 300 seconds how much power is needed (joules) ?
Help me understand the mathematical model that will show the relationship between this? what about inserting an "R value?"
Google searching brings up this often.
The Specific Heat Capacity equation is:
c = ΔQ / (m
I will admit word problems are my worst when dabbling in math so I need a little help.
Lets assume I have a 5000 mm volume of water resting at 25 c. I want to rise the temperature to 100 c in 300 seconds how much power is needed (joules) ?
Help me understand the mathematical model that will show the relationship between this? what about inserting an "R value?"
Google searching brings up this often.
The Specific Heat Capacity equation is:
c = ΔQ / (m
Comments
The joule is a measure of energy, not power. Power will be the rate you have to add energy (watts = joules/second). You have 300 seconds to make the 75°C change.
You are close, but you need to be clear about the units:
4.186joule/gram c = ΔQ / (5000ml × 75°C)
Solve for ΔQ and divide by time and you will have the power in watts. That of course assumes there is no loss, an R value of infinity, a good thermos bottle. The equations become more complicated when there is loss, because the loss rate (in watts, joules per second) is proportional to the temperature difference across the insulation.
Thank you for the quick reply, now about the complicated part.
Could I approximate and still come close?
Consider this - calculating as above if I can get my units straight. Then reduce by an efficiency percentage? In other words answer - 30%?
Would this get me close or is it too sloppy?
This is a big "it depends".
What kind of container are you using?
I recall in some of my chemistry labs we used two styrofoam coffee cups as our calorimeter. It worked pretty well. There are additional tricks you can do to increase accuracy if you plot the temperature over time. By taking into account how fast the cups would lose heat, we could adjust our measurements. These adjusted measurements were pretty darn close to the theoretical values. (We were testing the heat given off in exothermic reactions.)
How are you adding the heat?
The more information you can give the better answers we can give. (Though I think Tracy manages to give good answers all the time.)
Edit: I should have mentioned that the point of the coffee cup story is there are situations when the theoretical closely match real world experience but there are also going to be plenty of times when the real world isn't going to be anywhere near what would be expected in ideal circumstances. I'm sure there are many containers and heating element combinations where the ideal computed energy to heat the material wouldn't be nearly enough to compensate for inefficient heat transfer and heat losses to the environment. There will be times when your "30%" correction would work reasonably well but there will also be times with the correction should be more like 300%. The guessing could be reduced with a few careful measurements. How fast does the temperature of water really change when using a known amount of power? You'd probably want to take these sorts of measurements at both of the expected extremes (lowest expected temperature vs highest expected temperature). My guess is the heat transfer will be more efficient with low temperature water.
Just to clarify the words of the question
and to repeat what Tracy Allen said, the words should really be
"how much energy is needed (joules)?"
or
"how much power is needed (watts)?"
You can give an answer to both these values.
In terms of heat loss, yes you probably can approximate. The volume of water, the heating element, the time to heat and the container are rather similar to a large coffee urn. It is heating up so quickly that it might only be a few percentage points of heat loss. Much less if the container is insulated.
I like to use two simple formulas:
1) 1 watt is 1 joule per second. W = J/S
2) 4.186 joules will raise 1ml of water 1 degree C
You can mix and match these, by then saying that 4.186 watts will raise 1ml of water 1 degree per second. So 4.186*5000 watts will raise 5000ml of water 1 degree per second. That will take 75 seconds to raise the water from 25C to 100C. But you are doing it over 300 seconds, so it will be less watts.
You can get a feel for the numbers and make sure you are not out by 1000, which can be easy to do when you are given formulas like