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victus00 [196]
1 year ago
14

If the temperature rises by 9. 9 degrees, what is the corresponding temperature increase in degrees celsius?

Physics
1 answer:
Yanka [14]1 year ago
7 0

Main Answer:

Assume that the temperature rises by 9.9 F

Initial Temperatures will be C1, F1

Final temperatures will be C2, F1 + 9.9

The relation between celsius and fahrenheit temperature is as follows:

C = 5/9(F - 32)

Increase in temperature in celsius will be

C2 - C1 = F2 - F1

C2 - C1 = 5/9 ( F1 +9.9 -32) - 5/9 ( F1 -32)

C2 - C1 = 5/9 ( F1 +9.9 -32 - F1 +32)

C2 - C1 = 5/9 (9.9)

C2 - C1 = 5.5C

Explanation:

What is temperature?

Temperature is defined as the degree of coldness or hotness of an object or a substance. Generally, it is measured in various temperature scales. They are celsius, fahrenheit, kelvin.

To know more about temperature, please visit:

brainly.com/question/25677592

#SPJ4

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Answer:

steady state temperature =88.7deg C

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Explanation:

A 1 m long wire of diameter 1mm is submerged in an oil bath of temperature 25-degC. The wire has an electrical resistance per unit length of 0.01 Ω/m. If a current of 100 A flows through the wire and the convection coefficient is 500W/m2K, what is the steady state temperature of the wire? From the time the current is applied, how long does it take for the wire to reach a temperature within 1-degC of the steady state value? The density of the wire is 8,000kg/m3, its heat capacity is 500 J/kgK and its thermal condu

The diameter of the wire is known to be=1mm

properties=

The density of the wire is 8,000 kg/m3,

heat capacity is 500 J/kgK

themal conductivity is 20W/m.K

electrical resistance per unit length of 0.01 Ω/m

from lump capavity method

B_{i} =\frac{hr/2}{k}

500*(2.5*10^-4)/20

0.006<0.1

we know also, to find steady state temperature

\piDh(T-Tinf)=I^{2} R_{e}

make T the subject of the equation , we have

T=25+\frac{100^2*0.01}{\pi*0.001*500 }

T=88.7 degC

rate of chnage in temperature

dT/dt=\frac{I^2*Re}{rho*c*\pi*D^2/4 } -\frac{4h}{rho*c*D} (T-Tinf)

at t=o and integrating both sides\frac{T-Tinf-(I^2*Re/\pi*Dh) }{Ti-Tinf-(I^2*Re/\pi*Dh } =exp\frac{-4ht}{rho*c*D}

we have

\frac{87.7-25-63.7}{25-25-63.7} =exp\frac{4*500t}{8000*500*0.001}

t=8.31s

steady state temperature =88.7deg C

t=time within  1 degC of it steady stae is 8.31s

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