Start Free Mock LoginAlready registered? Login here

Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoHeat TransferMedium2 min

PhysicsMediumnumerical

A slab has two layers in series: 5.0 cm of material of conductivity 200 W m⁻¹ K⁻¹ and 3.0 cm of material of conductivity 50 W m⁻¹ K⁻¹. The outer faces are at 120°C and 20°C. Find the temperature of the interface, in °C.

Answer:

90.60

Solution:

$R_{1} = \frac{0.05}{200} = 2.50 \times 1 0^{- 4}$ , $R_{2} = \frac{0.03}{50} = 6.00 \times 1 0^{- 4}$ . $q = \frac{100}{R _{1} + R _{2}} = 117647$ W/m². Interface = $120 - q R_{1} = 90.6^{\circ}$ C.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Heat Transfer

⏱ 2m

Doubts & Discussion

Loading discussions...