Water temperature calculations

[gshock]

If 45lbs of 45F water plus 7lbs of 85F equals 52lbs and 51F water. How much lbs of 45F and 85F do you need to get 52lbs of 65F water and what is the formula for achieving it?

[swansont]

If 45lbs of 45F water plus 7lbs of 85F equals 52lbs and 51F water. How much lbs of 45F and 85F do you need to get 52lbs of 65F water and what is the formula for achieving it?

 

You should be able to figure out the formula from the given information; there's some quantity which depends on how much water you have and what the temperature is.

 

The actual formula is Q = cmT, where Q is the energy content, c is the specific heat capacity, m is mass and T is temperature. Energy is a conserved quantity. You should be able to use this to figure out the formula you need to solve the problem.

[Bignose]

while the amount of energy in water isn't exactly constant with temperature (i.e. the heat capacity of water is a function of temperature), for all but the most accurate calculations, it is fine to assume that the heat capacity of water is a constant.

[gshock]

You should be able to figure out the formula from the given information; there's some quantity which depends on how much water you have and what the temperature is.

 

The actual formula is Q = cmT, where Q is the energy content, c is the specific heat capacity, m is mass and T is temperature. Energy is a conserved quantity. You should be able to use this to figure out the formula you need to solve the problem.

 

 

For clarity and for me to get a full understanding please give me an example by using the numbers I posted.

[swansont]

For clarity and for me to get a full understanding please give me an example by using the numbers I posted.

 

That defeats the purpose of working the problem. How about you give it a try, and let people correct you if you get it wrong.

[swansont]

We want to conserve energy, and we have Q=cmT, which we will rewrite as Q/c = mT, since the specific heat capacity is a constant for this problem. We're OK using F, since the equation is really for changes in energy and temperature

 

45 lbs of 45 F water = 2025 lb-F (which are horrendous units, but as long as we're consistent, it should work out)

7 lbs of 85F water = 595 lb-F

 

Which totals 2620 lb-F for Q/c, and the total weight is 52 lbs. Applying the equation to get the mass (weight) we get 50.4 F That's how the setup problem was done.

[gshock]

We want to conserve energy, and we have Q=cmT, which we will rewrite as Q/c = mT, since the specific heat capacity is a constant for this problem. We're OK using F, since the equation is really for changes in energy and temperature

 

45 lbs of 45 F water = 2025 lb-F (which are horrendous units, but as long as we're consistent, it should work out)

7 lbs of 85F water = 595 lb-F

 

Which totals 2620 lb-F for Q/c, and the total weight is 52 lbs. Applying the equation to get the mass (weight) we get 50.4 F That's how the setup problem was done.

 

 

So if I wanted to get 65F of 52lbs of water do this - 65X52 = 3380 since its two temperature of water do I then do 3380/2=1690/85 = 19.88lbs. 1690/45=37.55lbs. 37.55+19.88= 57.43lbs. It doesn't add up to 52lbs of water, where did I go wrong?

[Mr Skeptic]

I suggest you do this problem in two steps: first you find the ratio of water to make your desired temperature. Then use that ratio to find out how much water specifically to get the desired weight. For the first ratio I suggest adding 1 pound of the cooler water with n pounds of the warmer water, then solve for n.

[swansont]

you want x amount of one temperature, and 52-x of the other

 

45x + (52-x)85 = 52*65

 

40x = 1040

 

You get 26 lbs (which, since 65 is midway between 45 and 85, should seem reasonable — you want equal amounts)

 

I'm sorry if this process seemed uncooperative; we get a lot of people here with homework questions, and that's what this looked like. If you'd made it known that this was a real live pizza-related problem, this would have gone faster

[gshock]

is the formula 45+85= 130 then 52*65=3380/130=26, 52-26 = 26. 85-45=40. Whats the purpose of 40 or 1040?

Edited August 7, 2010 by gshock

[swansont]

40 is the difference in temperature between hot and cold. 1040 is specific to this problem. I PM-ed the general formula

[gshock]

40 is the difference in temperature between hot and cold. 1040 is specific to this problem. I PM-ed the general formula

 

 

Now I'm even more confuse, I thought I finally had the answer for the first equation with 45+85=130,52*65=3380/130=26,52-26=26; so 26lbs at 45F and 26lbs at 85F.In your post you said 1040 was specific I don't understand what that means. Did you multiply 26 by 40 to get 1040? Why didn't you do (1040-52*T)/40 in the first equation like you did in the second? I don't understand how you arrive at 4420 for the second equation. When you wrote 45x+85(52-x), does that mean 45 multiplied by x plus 85 multiplied by 52 minus x. Your formula still doesn't show how you went about getting 32.5lbs for 45 and 19.5 for 85. Wouldn't my equation work? If your tired of me then I understand but I cannot let it go until I understand fully. If you reply please try to explain has simply as you can.

 

Now I'm even more confuse, I thought I finally had the answer for the first equation with 45+85=130,52*65=3380/130=26,52-26=26; so 26lbs at 45F and 26lbs at 85F.In your post you said 1040 was specific I don't understand what that means. Did you multiply 26 by 40 to get 1040? Why didn't you do (1040-52*T)/40 in the first equation like you did in the second? I don't understand how you arrive at 4420 for the second equation. When you wrote 45x+85(52-x), does that mean 45 multiplied by x plus 85 multiplied by 52 minus x. Your formula still doesn't show how you went about getting 32.5lbs for 45 and 19.5 for 85. Wouldn't my equation work? If your tired of me then I understand but I cannot let it go until I understand fully. If you reply please try to explain has simply as you can.

 

 

I give up I think this goes beyond my IQ.

[Guest Girish]

Hi,

I believe part of the difference is that BeerSmith accounts for differences in mash volume. For example, if you have a 15 gallon sankey keg for a mash tun but you are using only 5 gallons of it for the mash then you do not necessarily heat the entire mash tun to 150F. BeerSmith attempts to compensate by only including a portion of the equipment effect proportional to the volume used. That is why your 20lb and 10lb examples have similar strike temperatures.

 

The mash tool does not have separate settings for mash tun specific heat and some starting temperatures, which is why the recipe calculator is slightly more accurate. It assumes a temperature near room temperature, and moderate specific heat. In cases like the second one which you illustrate this can have a significant effect - especially for a large mash tun that is cold.

 

Cheers,