The temperature dropped from 75 degrees to 50 degrees. What was the percent decrease in the temperature?

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Answer 1 · 2016-05-10T10:32:23

#33 1/3 %# as an exact value

#color(blue)("Shortcut method")#

#(75-50)/75xx100 =color(blue)(33.3bar3%)#

The #bar3# means that it goes on repeating for ever.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("First principles method")#

Percentage is a fraction where the bottom number (denominator) is fixed at 100.

We need to change #25/75# into #("some value")/100#

Let the unknown value be #x#

#=>25/75-=x/100#

Find some way to convert 75 into 100, do the same thing to the 25 and we have our answer.

If we divide 75 by 75 we have 1. Then if we multiply the 1 by 100 we have our 100

#75-:75xx100 -> 75xx100/75 =100#

Do the same thing to the top (numerator)

#25xx100/75 = 33 1/3#

So #25/100-=(33 1/3)/100 ->color(blue)( 33 1/3%)#

Answer 2 · 2016-05-10T11:04:43

It is notcorrect to talk about percent of temperature in an arbritary scale like the centrigrade scale.

The only meaningfull is the absolute scale.

So 75 °C corresponds to 75+273.15=348.15K,

and 50°C corressponds to 50+273.15=323.15K.

The percental decrease will be

#100xx25/348.15=7.2%#.

In close vessel taking the law:

#P_1/P_2=T_1/T_2# will be accompanied by a 7.2% decrease of pressure.