Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
If absolute zero is -273. 15°C on Celsius temperature scale, then the absolute zero on the Fahrenheit scale is
Q. If absolute zero is -273. 15
∘
C on Celsius temperature scale, then the absolute zero on the Fahrenheit scale is
3038
194
Thermal Properties of Matter
Report Error
A
-227.15
∘
F
14%
B
-453.15
∘
F
16%
C
-459.67
∘
F
60%
D
-49l.67
∘
F
9%
Solution:
Celsius and Kelvin scale are related as
5
C
=
9
F
−
32
or
5
−
273.15
=
9
F
−
32
⇒
F
=
−
459.6
7
∘
F