Chemistry, asked by shreyas37991, 11 months ago

F 9 5 c 32 is there a temperature which is numerically same in fahrenhrit and celsius

Answers

Answered by ItsSpiderman44
0

Answer:

Method to draw the graph of linear equation in two variables:

1. Let the linear equation in two variables be ax+by+c=0

Write the linear equation and Express Y in terms of x.

Y=(-ax+c)/b......(1)

2.Put put arbitrary value of x in equation 1 and find the corresponding values of y.

3.Form a table by writing the value’s of y to the corresponding values of x.

4.Draw the coordinates axes on graph paper and take a suitable scale to plot points from the table on graph paper.

5.Join the points and we get a straight line and produced it on both sides.

Hence, the straight line so obtained is the required graph of given linear equation.

It is enough to plot two points Corresponding to two solutions and join them by a line.

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Solution:

Given, linear equation in Fahrenheit and Celsius is

F=(9/5)C +32

5F-9C =160.........(1)

When C=0, then F= 160/5=32

When F= 0, then C = -160/9= -17.8

[Table and graph are on the attachment ]

ii)

If temperature is 30°C i.e C= 30°C

Then from equation 1 we get,

5F-9C=160

5F – 9×30=160

5F-270= 160

5F= 160+270

5F= 430

F=430/5=86°F

Temperature in Fahrenheit= 86°F

iii)

If temperature is 95° F , F= 95°F

Then from equation 1 ,

5F-9C=160

5×95 – 9C= 160

475-9C = 160

475-160= 9C

315= 9C

C= 315/9= 35°

Temperature in Celsius = 35°C

iv)

If the temperature is 0°C

Then, from eq 1

5F-9C=160

5F-9×0=160

5F=160

F=160/5=32

Temperature in Fahrenheit = 32°F

If the temperature is 0°F

Then, from eq 1

5F-9C=160

5×0-9×C=160

-9C=160

C=-160/9= -17.8°C

Temperature in Celsius= -17.8°C

v)

Yes, if we take both temperature are equal, i.e, C=F

Now ,from eq 1 we get

5F-9C=160

5F= 9C+160

5F= 9F+160

5F-9F=160

-4F= 160

F= -160/4= -40

Hence, F=C=-40°

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