Question

In a random temperature scale X, water boils at 200°c & freezes at 20°X. Find the boiling point of the liquid in this scale if it boils at 62°c​

Answer 1

Answer:

Boiling point of the liquid in new scale is 131.6 degree X

Explanation:

As we know that temperature scale is always linear scale

So here we can say that

$\frac{T_x - 20}{200 - 20} = \frac{T_c - 0}{100 - 0}$

so we have

$\frac{T_x - 20}{180} = \frac{T_c}{100}$

now we know that liquid boils at 62 degree C

so here we will have

$T_x - 20 = 1.80 T_c$

here we have

$T_c = 62$

$T_x = 131.6 ^oX$

#Learn

Topic : Temperature scale

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Answer 2

Explanation:

As we know that temperature scale is always linear scale

So here we can say that

\frac{T_x - 20}{200 - 20} = \frac{T_c - 0}{100 - 0}

200−20

T

x

−20

=

100−0

T

c

−0

so we have

\frac{T_x - 20}{180} = \frac{T_c}{100}

180

T

x

−20

=

100

T

c

now we know that liquid boils at 62 degree C

so here we will have

T_x - 20 = 1.80 T_cT

x

−20=1.80T

c

here we have

T_c = 62T

c

=62

T_x = 131.6 ^oXT

x

=131.6

o

X