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In summary, the conversation discusses the temperature changes of a can of varnish that was brought into a room with a controlled temperature of 74 deg F from a storage temperature of 40 deg F. After 2 hours, the temperature of the varnish rose to 58 deg F, but for best results, it should be used at 68 deg F. The question is how much longer do they have to wait to use the varnish and get the best results. The person asking the question is in precalculus and has attempted to solve the problem using different methods but is unsure if they are correct.
  • #1
wegnes
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  • #2
a can of varnish that was stored @ 40 deg F was brought in a living room where the temp is controlled at a constant 74 deg F. After 2 hours the Temp of the varnish has risen to 58 deg F. For best results the ideal Temp to use the varnish is 68 deg F. How much longer do we have to wait to use the varnish and get best results?
Please help!
 
  • #3
Help by telling you the answer? What have you done on this yourself?
 
  • #4
A few questions for you:

1) What level of study are you at? (This will help the community identify what is likely the desired answer)

2) Were you given any other information in this problem?The reason I ask is that, if one assumes that temperature change in this problem is linear, then the problem is straightforward. If you've learned calculus, this changes things.
 
  • #5
I am in precalculus, here's what I tried...

BrendanH said:
A few questions for you:

1) What level of study are you at? (This will help the community identify what is likely the desired answer)

2) Were you given any other information in this problem?


The reason I ask is that, if one assumes that temperature change in this problem is linear, then the problem is straightforward. If you've learned calculus, this changes things.


I tried 2 hr/18 degrees x 10 degress = 1.11
 
  • #6
Regarding the story problem ques:

I am new to this website. I apologize for not showing my work. I have worked on it a lot actually I just don't know if I'm right or not.

I tried...
2 hr/ 18 degrees x 10 degrees = degrees cancel so 1.11 hour

I tried also with the Newtons law of heating and cooling equation: but not sure if the K value I come up with using t= 2 hours and the temp difference 40 to 58 it the one I should use to find the time when the temp changes 10 more degrees.
 
  • #7
I have no idea what you mean by "I tried...2 hr/ 18 degrees x 10 degrees = degrees cancel so 1.11 hour". Surely you don't mean you just put numbers together randomly!

Newton's law of heating and cooling applies here. Please show us exactly what you did.
 

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