Question

At 160°C, the resistance of WIRE IS 30 ohm WHEN THE WIRE Is PLACED IN A LIQUID BATH THE R DECREASES TO 20 OHM CALCULATE the TEMPERATURE of bath . If alpha = 3.9 x 10 -3 C ​

Answer 1

Answer:

Given : R

T

=60 ohms @ T=500

o

C

Also, R

T

′

=20 ohms @ T

′

=20

o

C

Resistance at a given temperature R

T

=R

T

′

[1+α(T−T

′

)]

∴ 60=20[1+α(500−20)]

⟹ α=

480

2

=0.00417

Let resistance of wire is 25 ohm at temperature T

′′

.

∴ R

T

′′

=R

T

′

[1+α(T

′′

−T

′

)]

OR 25=20[1+(0.00417)(T

′′

−20)]

⟹ T

′′

=80

o

C

Answer 2

Given : R

Given : R T

Given : R T

Given : R T =60 ohms @ T=500

Given : R T =60 ohms @ T=500 o

Given : R T =60 ohms @ T=500 o C

Given : R T =60 ohms @ T=500 o CAlso, R

Given : R T =60 ohms @ T=500 o CAlso, R T

Given : R T =60 ohms @ T=500 o CAlso, R T ′

Given : R T =60 ohms @ T=500 o CAlso, R T ′

Given : R T =60 ohms @ T=500 o CAlso, R T ′

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o C

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′ [1+α(T−T

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′ [1+α(T−T ′

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′ [1+α(T−T ′ )]

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′ [1+α(T−T ′ )]∴ 60=20[1+α(500−20)]

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′ [1+α(T−T ′ )]∴ 60=20[1+α(500−20)]⟹ α=

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′ [1+α(T−T ′ )]∴ 60=20[1+α(500−20)]⟹ α= 480

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′ [1+α(T−T ′ )]∴ 60=20[1+α(500−20)]⟹ α= 4802

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′ [1+α(T−T ′ )]∴ 60=20[1+α(500−20)]⟹ α= 4802

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′ [1+α(T−T ′ )]∴ 60=20[1+α(500−20)]⟹ α= 4802 =0.00417

Given : R T =60 ohms @ T=500 o CAlso, R T ′ =20 ohms @ T ′ =20 o CResistance at a given temperature R T =R T ′ [1+α(T−T ′ )]∴ 60=20[1+α(500−20)]⟹ α= 4802 =0.00417Let resistance of wire is 25 ohm at temperature