Quiescent air is at 20°C (treat air as ideal gas). A plate of depth 50 cm and width 40 cm has a temperature of 40°C on the top surface. Calculate the heat dissipated from the top surface.

Explanation:

The heat dissipated or loss at the top of the surface is given by

Qloss = -KA ∆T

Qloss = -KA (Ts - Ta)

Where

K is thermal conductivity of material

A is the cross-sectional area

L is the thickness or length of material

Ta is the temperature of air.

Ts is the temperature at the surface

Ts–Ta is the difference between skin temperature and air temperature.

Given that,

Temperature of air Ta=20°C

Plate dimensions are 50cm by 40cm

Converting to metre = 0.5m by 0.4m

Area = length × breadth

Area=0.5×0.4

Area=0.2m²

The temperature at surface

Ts=40°C

Let assume an aluminium plate then, k = 0.0263 W/m⋅K,

We don't need to convert the temperature to Kelvin, since we are going to use the change in temp.

Applying the formulas

Qloss = -KA (Ts - Ta)

Qloss = - 0.0263 × 0.2 × (40-20)

Qloss= -0.0263 × 0.2 × 20

Qloss= -0.1052J

The heat loss is 0.1052J

adding a(n) to an electric circuit would increase the magnetic field around a wire;

b. solenoid

a solenoid is a type of electromagnet whose purpose is to generate a controlled magnetic field.

if the purpose of the solenoid is instead to impede changes in the electric current, a solenoid can be more specifically classified as an inductor rather than an electromagnet.