Given :
▪ Initial temp. = -10°C
▪ Final temp. = 35°C
To Find :
▪ The change in temp. on the fahrenheit scale.
Formula :
✒ If Tc and Tf are the temperatures of a body on celcius and fahrenheit, then,
\begin{lgathered}\bigstar\:\underline{\boxed{\bf{\pink{\dfrac{T_C-0}{100}=\dfrac{T_F-32}{212-32}}}}}\\ \\ \bigstar\:\underline{\boxed{\bf{\green{T_F=1.8T_C+32}}}}\end{lgathered}
★
100
T
C
−0
=
212−32
T
F
−32
★
T
F
=1.8T
C
+32
Calculation :
\begin{lgathered}\implies\sf\:(T_F)_1=1.8(T_C)_1+32\\ \\ \implies\sf\:(T_F)_1=1.8(-10)+32\\ \\ \implies\sf\:(T_F)_1 =-18+32\\ \\ \implies\bf\:\red{(T_F)_1=14\:F}\\ \\ \implies\sf\:(T_F)_2=1.8(T_C)_2+32\\ \\ \implies\sf\:(T_F)_2=1.8(35)+32\\ \\ \implies\sf\:(T_F)_2=63+32\\ \\ \implies\bf\:\blue{(T_F)_2=95\:F}\\ \\ \rightarrow\sf\:\Delta{T_F}=(T_F)_2-(T_F)_1\\ \\ \rightarrow\sf\:\Delta{T_F}=95-14\\ \\ \rightarrow\underline{\boxed{\bf{\purple{\Delta{T_F}=81\:F}}}}\end{lgathered}
⟹(T
F
)
1
=1.8(T
C
)
1
+32
⟹(T
F
)
1
=1.8(−10)+32
⟹(T
F
)
1
=−18+32
⟹(T
F
)
1
=14F
⟹(T
F
)
2
=1.8(T
C
)
2
+32
⟹(T
F
)
2
=1.8(35)+32
⟹(T
F
)
2
=63+32
⟹(T
F
)
2
=95F
→ΔT
F
=(T
F
)
2
−(T
F
)
1
→ΔT
F
=95−14
→
ΔT
F
=81F
Answers
Answered by
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I don't know this answer please sorry
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