SOLUTION: Hi, can you kindly please assist me with this question, thank you. Question 1 According to Mug & Bean barista, the optimum temperature to serve coffee is between 1550𝐹 and

Algebra ->  Test -> SOLUTION: Hi, can you kindly please assist me with this question, thank you. Question 1 According to Mug & Bean barista, the optimum temperature to serve coffee is between 1550𝐹 and      Log On


   

Question 1195999: Hi, can you kindly please assist me with this question, thank you.
Question 1
According to Mug & Bean barista, the optimum temperature to serve coffee is between
1550𝐹 and1750𝐹. Suppose coffee is poured at a temperature of 2000𝐹, and after 2 minutes
in a 700𝐹 room it has cooled to1800𝐹. When is the coffee first cooled enough to serve?
When is the coffee too cold to serve? Round your answer to the nearest half minute.
[15 marks]
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39625) About Me  (Show Source):
You can put this solution on YOUR website!
None of this makes sense:
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the optimum temperature to serve coffee is between
1550𝐹 and1750𝐹. Suppose coffee is poured at a temperature of 2000𝐹, and after 2 minutes
in a 700𝐹 room it has cooled to1800𝐹.
------------------------------------------------------------------------

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

Hi, can you kindly please assist me with this question, thank you.
Question 1
According to Mug & Bean barista, the optimum temperature to serve coffee is between
1550𝐹 and1750𝐹. Suppose coffee is poured at a temperature of 2000𝐹, and after 2 minutes
in a 700𝐹 room it has cooled to1800𝐹. When is the coffee first cooled enough to serve?
When is the coffee too cold to serve? Round your answer to the nearest half minute.
[15 marks]
ALL of your temperatures are OFF. I'd assume that you copied and pasted this problem and didn't bother to check to see
if it was pasted correctly. As such, 1550𝐹 and 1750𝐹 s/b 155oF and 175oF
                                     2000𝐹 s/b 200oF
                                     700𝐹 room s/b 70oF room
                                     1800𝐹 s/b 180oF
You can use Newton's Law of Cooling.......any of the following 3 formulae should work: 

I usually use the 1st formula, And, as we need to first find the cooling rate, we have the following:
       , where: t = time taken to get to a COOLED temperature (2 minutes, in this case)
                                             T%28t%29 = TEMPERATURE (T) at a given time (t)___(180oF, in this case) 
                                              T%5Bs%5D = SURROUNDING Temperature (70oF, in this case)
                                              T%5Bo%5D = ORIGINAL/INITIAL temperature (200oF, in this case)
                                               k = the CONSTANT or COOLING rate (UNKNOWN, in this case)

We get: matrix%281%2C3%2C+180%2C+%22=%22%2C+70+%2B+%28200+-+70%29+%2A+e%5E%28-+2k%29%29 ----- Substituting 180o for T(t), 70o for T%5Bs%5D, 200o for T%5Bo%5D, and 2 for t.
        
       matrix%281%2C3%2C+-+2k%2C+%22=%22%2C+ln+%2811%2F13%29%29 ----- Converting to NATURAL LOGARITHMIC (ln) form
          matrix%281%2C5%2C+k%2C+%22=%22%2C+ln%2811%2F13%29%2F%28-+2%29%2C+%22=%22%2C+0.083527042%29

As the optimum temperature to serve coffee is between 155o and 175o, then at 175o it should be cool enough to be served.
   We then get: 
                  matrix%281%2C3%2C+175%2C+%22=%22%2C+70+%2B+%28200+-+70%29+%2A+e%5E%28-+.083527042t%29%29 ----- Substituting 175o for T(t), 70o for T%5Bs%5D, 200o for T%5Bo%5D, and .083527042 for k.
                  
                  matrix%281%2C3%2C+-.083527042t%2C+%22=%22%2C+ln+%2821%2F26%29%29 ----- Converting to NATURAL LOGARITHMIC (ln) form
Time at which the coffee has cooled enough to serve. or
==================================
As the optimum temperature to serve coffee is between 155o and 175o, then a temperature that's less than 155o makes
it too cold to serve.

              We then get: 
                  matrix%281%2C3%2C+70+%2B+%28200+-+70%29+%2A+e%5E%28-+.083527042t%29%2C+%22%3C%22%2C+155%29 ----- Substituting 155o for T(t), 70o for T%5Bs%5D, 200o for T%5Bo%5D, and .083527042 for k.
                        
                                  matrix%281%2C3%2C+-.083527042t%2C+%22%3C%22%2C+ln+%2817%2F26%29%29 ----- Converting to NATURAL LOGARITHMIC (ln) form
                                               

Time at which the coffee has become too cold to serve. or