Given,
emissivity(Ξ΅) = 1 (since we are not given),
Stefan-Boltzmann constant(Ο )= 6 Γ 10β»βΈ W/mΒ²-Kβ΄,
Surface area of cylindrical wall heater(A) = 2Οrh where
radius of wall heater(r) = 6 mm = 6 Γ 10β»Β³ m and
length of heater(h) = 0.6 m, and
TemperatureΒ ofΒ heater(T)=?
Since P = Ξ΅ΟATβ΄
P = Ξ΅Ο(2Οrh)Tβ΄
Making T subject of the formula, we have
Temperature of heater
T = β΄β[P/Ξ΅Ο(2Οrh)]
Since P = 1.5 kW = 1.5 Γ 10Β³ W
Substituting the values of the variables into the equation, we have
T = β΄β[P/Ξ΅Ο(2Οrh)]
T = β΄β[1.5 Γ 10Β³ W/(1 Γ 6 Γ 10β»βΈ W/mΒ²-Kβ΄ Γ 2Ο Γ 6 Γ 10β»Β³ m Γ 0.6 m)]
T = β΄β[1.5 Γ 10Β³ W/(43.2Ο Β Γ 10β»ΒΉΒΉ W/Kβ΄)]
T = β΄β[1.5 Γ 10Β³ W/135.72 Β Γ 10β»ΒΉΒΉ W/Kβ΄)]
T = β΄β[0.01105 Γ 10ΒΉβ΄ Kβ΄)]
T = β΄β[1.105 Γ 10ΒΉΒ² Kβ΄)]
T = 1.0253 Γ 10Β³ K
T = 1025.3 K
Hence, the correct answer is option c. 1025K
Given,
emissivity(Ξ΅) = 1 (since we are not given),
Stefan-Boltzmann constant(Ο )= 6 Γ 10β»βΈ W/mΒ²-Kβ΄,
Surface area of cylindrical wall heater(A) = 2Οrh where
radius of wall heater(r) = 6 mm = 6 Γ 10β»Β³ m and
length of heater(h) = 0.6 m, and
TemperatureΒ ofΒ heater(T)=?
Since P = Ξ΅ΟATβ΄
P = Ξ΅Ο(2Οrh)Tβ΄
Making T subject of the formula, we have
Temperature of heater
T = β΄β[P/Ξ΅Ο(2Οrh)]
Since P = 1.5 kW = 1.5 Γ 10Β³ W
Substituting the values of the variables into the equation, we have
T = β΄β[P/Ξ΅Ο(2Οrh)]
T = β΄β[1.5 Γ 10Β³ W/(1 Γ 6 Γ 10β»βΈ W/mΒ²-Kβ΄ Γ 2Ο Γ 6 Γ 10β»Β³ m Γ 0.6 m)]
T = β΄β[1.5 Γ 10Β³ W/(43.2Ο Β Γ 10β»ΒΉΒΉ W/Kβ΄)]
T = β΄β[1.5 Γ 10Β³ W/135.72 Β Γ 10β»ΒΉΒΉ W/Kβ΄)]
T = β΄β[0.01105 Γ 10ΒΉβ΄ Kβ΄)]
T = β΄β[1.105 Γ 10ΒΉΒ² Kβ΄)]
T = 1.0253 Γ 10Β³ K
T = 1025.3 K
Hence, the correct answer is option c. 1025K