3. Explain the construc labelled diagram. ) . (1) 150 °F $(a) Convert the given temperatures to °C. (F-32]x [f5 (b) Convert the given temperatures to °F. [CX-9]+32 (ii) 48 °F (ii) 32 °C (i) 150 °C
Answers
Explanation:
C=
9
5F−160
(i) Putting F=86
∘
, we get C=
9
5(86)−160
=
9
430−160
=
9
270
=30
∘
Hence, the temperature in Celsius in 30
∘
C.
(ii) Putting C = 35
∘
, we get 35
∘
=
9
5(F)−160
⇒315
∘
=5F−160
⇒5F=315+160=475
∴F=
5
475
=95
∘
Hence, the temperature in Fahrenheit is 95
∘
F.
(iii) Putting C = 0
∘
, we get
0=
9
5F−160
⇒0=5F−160
⇒5F=160
∴F=
5
160
=32
∘
Now, putting F = 0
∘
, we get
C=
9
5F−160
⇒C=
9
5(0)−160
=(−
9
160
)
∘
If the temperature is 0
∘
C, the temperature in Fahrenheit is 32
∘
and if the temperature is 0
∘
F, then the temperature in Celsius is (−
9
160
)
∘
C.
(iv) Putting C=F, in the given relation, we get
F=
9
5F−160
⇒9F=5F−160
⇒4F=−160
∴F=
4
−160
=−40
∘
Hence, the numerical value of the temperature which is same in both the sacales is −40.
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Answer:
C=
9
5F−160
(i) Putting F=86
∘
, we get C=
9
5(86)−160
=
9
430−160
=
9
270
=30
∘
Hence, the temperature in Celsius in 30
∘
C.
(ii) Putting C = 35
∘
, we get 35
∘
=
9
5(F)−160
⇒315
∘
=5F−160
⇒5F=315+160=475
∴F=
5
475
=95
∘
Hence, the temperature in Fahrenheit is 95
∘
F.
(iii) Putting C = 0
∘
, we get
0=
9
5F−160
⇒0=5F−160
⇒5F=160
∴F=
5
160
=32
∘
Now, putting F = 0
∘
, we get
C=
9
5F−160
⇒C=
9
5(0)−160
=(−
9
160
)
∘
If the temperature is 0
∘
C, the temperature in Fahrenheit is 32
∘
and if the temperature is 0
∘
F, then the temperature in Celsius is (−
9
160
)
∘
C.
(iv) Putting C=F, in the given relation, we get
F=
9
5F−160
⇒9F=5F−160
⇒4F=−160
∴F=
4
−160
=−40
∘
Hence, the numerical value of the temperature which is same in both the sacales is −40.
Explanation:
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