My Math Forum Water temperature calculation

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 August 4th, 2010, 09:50 AM #1 Newbie   Joined: Aug 2010 Posts: 4 Thanks: 0 Water temperature calculation If 45lbs of 45F water plus 7lbs of 85F water can give you 52lbs of 51F water. How much pounds of the 45F plus the 85F water do you need to get 52lbs of 65F water? What is the formula for achieving the answer and please give an example by using the figures?
 August 4th, 2010, 12:32 PM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Water temperature calculation Hello doughboy33, That would be $\frac{45\, \mathrm{lbs} \cdot 45F+\7\, \mathrm{lbs} \cdot 85 F}{(45+7)lbs}= \frac{665F}{13}\approx50.3846153846153846153846153 8461$ How would you solve this: John played a game. 3 times, when he played, he got himself 5 points. 4 times when he played, he got himself 12 points. What was the average score of all games? Have you got an idea for a formula?
 August 6th, 2010, 09:39 AM #3 Newbie   Joined: Aug 2010 Posts: 4 Thanks: 0 Re: Water temperature calculation That still doesn't answer my question on how much of the 85F and 45F water you need to get 65F 52lbs water. Here's the answer to yours though 5+12/3+4=2.4points
 August 6th, 2010, 10:05 AM #4 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Water temperature calculation I meant something else, but didn´t formulate the question well. Your interpretation is a correct method. Do notice that you have to use brackets and that the answer is not exactly 2.4. So (5+12)/(3+4)=17/7$\approx$ 2.4. For your other question: let us say that we have x lbs water of 85F, and y lbs water of 45 F. That is $\frac{y\; \mathrm{lbs}\; \cdot 45F\,+\,x \;\mathrm{lbs} \cdot \;85F}{52lbs}=65F$ In the end, you have 52 lbs water (of 65 F). So x+y=52==>y=52-x. Substitute in the equation: $\frac{(52-x)\; \mathrm{lbs}\; \cdot 45F\,+\,x \;\mathrm{lbs} \cdot \;85F}{52lbs}=65F$ Now, solve for x. Can you go from here? Hoempa
 August 6th, 2010, 12:40 PM #5 Newbie   Joined: Aug 2010 Posts: 4 Thanks: 0 Re: Water temperature calculation Still confuse?
 August 6th, 2010, 12:44 PM #6 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Water temperature calculation Because ...? What is it that confuses you? Hoempa
 August 6th, 2010, 02:25 PM #7 Newbie   Joined: Aug 2010 Posts: 4 Thanks: 0 Re: Water temperature calculation Can you help me solve this problem its work related? I'm manager of a local pizza place my job is to produce the dough, but the machine that controls the flow and temperature of water is out of operation. We're currently mixing the water manually to maintain proofing consistency for every dough batch. It's a headache guessing how much hot and cold water you need to produce the dough. Can you just give me a straight answer avoiding the use of algebra and scientific terms? I'm more familiar with plus,multiplication, division and subtraction because I'm no math wiz. Please reply asap and post it in the forum. If you use the figures I gave you and solve the problem I can use that for calculations. The question is we use 52lbs of water per batch, if its 52lbs and we need it at 65F and we have both cold (45F) and warm (65F)water present; how much poundage of each water do I need to get 52lbs of 65F water? Please provide the equation without letters and show clearly how you arrive at the answer.
 August 7th, 2010, 05:56 AM #8 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Water temperature calculation I will give you the solution and a formula you can use for any temperature (in F) of water and any quantity (in lbs) of water. $\frac{(52-x)\; \mathrm{lbs}\; \cdot 45F\,+\,x \;\mathrm{lbs} \cdot \;85F}{52lbs}=65F$ You can safely eliminate the units, to only work with numbers: \frac{((52-x)45+85x}{52}=65 \frac{52 \cdot 45-45x+85x}{52}=65 2340+40x=3380 40x=1040 x=26 You need 52-26=26 lbs of 45F-water and 26 lbs of 85F-water to get 52 lbs of 65F-water. Assume you have a lbs of bF water, and c lbs of dF water, and you want f lbs of gF water. a+c=f. a=f-c $\frac{(f-c)lbs\; of\; bF-water+c \;lbs \;of\; dF \;water}{f\, lbs}=gF$ Remove the units: $\frac{b(f-c)+cd}{f}=g.$ I hope, this helps you, and your dough will work out well. Hoempa

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