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Fot an experiment, a liquid is kept in cold storage at a temperature of 20 degrees Fahrenheit(°F). The liquid is removed from cold storage and is warmed at a constant rate of 3°F per hour. Which of the following represents the temperature of the liquid, in degrees Celsius(°C), t hours after it has been removed from cold storage ?

(Note:°C=5/9(°F-32).) 

 

A)5/3t-20/3

B)5/3t-188/9

C)5/9t-20/3

D)27/5t+68

E)3t+20

 Jan 3, 2016
edited by Solveit  Jan 3, 2016

Best Answer 

 #6
avatar+111325 
+10

The temperature in F after t hours = 3t + 20

 

F →  C    ..... C =   (F - 32)/ (9/5)

 

So ....the temperature in  C  after t hours   = 

 

( 3t + 20 - 32)/ (9/5)  = ( 3t - 12) / (9/5)  = (5/9)(3t - 12)  = (5/9) * 3(t - 4)  =  (5/3)(t - 4)  =  (5/3)t - 20/3

 

 

 

cool cool cool

 Jan 3, 2016
 #1
avatar+2496 
+5

I found the answer (A)

 Jan 3, 2016
edited by Solveit  Jan 3, 2016
 #2
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0

E     E   E   EE   EEEE   E   'E'

 Jan 3, 2016
 #3
avatar+2496 
0

Why (E) ?

 Jan 3, 2016
 #4
avatar
+5

The sample STARTS at 20 degrees so the temperature is  20 + ?

Every hour it warms up an ADDITIONAL   3 degrees  so the temperature after  't' hours will be INCREASED  3 x t

It STARTED at 20 degrees    so  add both of the terms together to get the temperature of the sample  20 + (3 x t)

or  20 + 3t     Rewritten :  3t + 20

 

Now I re-read the question: it asks for Celcius

Convert to Celcius   5/9 (3t +20) -32 )

15/9t - 60/9

5/3t - 20/3

 

So YOU were RIGHT !   I was WRONG....sorry....I misread the question!!!

 Jan 3, 2016
edited by Guest  Jan 3, 2016
 #5
avatar+2496 
+5

Yea you right but you get it on °F but i need on °C so i put it into °C=5/9(°F-32) and get (A)

 Jan 3, 2016
 #6
avatar+111325 
+10
Best Answer

The temperature in F after t hours = 3t + 20

 

F →  C    ..... C =   (F - 32)/ (9/5)

 

So ....the temperature in  C  after t hours   = 

 

( 3t + 20 - 32)/ (9/5)  = ( 3t - 12) / (9/5)  = (5/9)(3t - 12)  = (5/9) * 3(t - 4)  =  (5/3)(t - 4)  =  (5/3)t - 20/3

 

 

 

cool cool cool

CPhill Jan 3, 2016

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