Convert Watts to Kelvins on Steel Plate: 850 Sunlight Spots

[douglasg14b]

How might i convert the temperature in kelvins of a 1/2inch 2" x 2" steel plate when i have 28322 watts of sunlight focused on it for 5min? also if i am stating this question wrong i will be focusing 850 2" x 2" spots of sunlight on a single 2" x 2" surface how might i calculate the temperature in K?

 

[vanesch]

The temperature will be the result of the balance between incoming power (your sunlight), and whatever it will lose (by conduction, radiation, reflection, ...).
So the incoming power is only one element. As the lost power is usually a strong function of the temperature (the higher the temperature, the more losses there are), we can write: W_out(T). So the end temperature will be the solution of W_in = W_out(T).

The hardest part is always to figure out W_out(T), that is: the lost power as a function of temperature.

 

[astrokreng]

First, we need to understand the relationship between watts and temperature. Watts are a unit of power, while temperature is a measure of the average kinetic energy of particles in a substance. The amount of power being applied to a substance can affect its temperature, but other factors such as the material's specific heat capacity and thermal conductivity also play a role.

To convert watts to kelvins, we need to know the specific heat capacity of steel and the thermal conductivity of steel. Let's assume that the steel plate is made of pure iron, which has a specific heat capacity of 0.45 joules per gram per kelvin and a thermal conductivity of 80.2 watts per meter per kelvin.

Next, we need to determine the amount of energy being absorbed by the steel plate. This can be calculated by multiplying the power (28322 watts) by the time (5 minutes or 300 seconds). This gives us a total energy of 8.4966 x 10^6 joules.

Using the specific heat capacity and thermal conductivity values, we can then calculate the change in temperature of the steel plate. The formula for this is ΔT = (Q/mc), where ΔT is the change in temperature, Q is the energy absorbed, m is the mass of the steel plate, and c is the specific heat capacity. Assuming a 2" x 2" steel plate with a thickness of 1/2 inch (or 0.0127 meters) and a density of 7.87 grams per cubic centimeter, we can calculate the mass of the steel plate to be 0.00023 kilograms.

Plugging in the values, we get ΔT = (8.4966 x 10^6 joules) / (0.00023 kg * 0.45 J/g*K) = 7.6 x 10^7 kelvins.

However, this calculation assumes that all of the energy from the sunlight is absorbed by the steel plate, which is not realistic. In reality, some of the energy will be reflected or transmitted. Additionally, the surface area of the steel plate may not be enough to evenly distribute the energy from 850 spots of sunlight. Therefore, the actual temperature of the steel plate will likely be lower than this calculated value.

To calculate the temperature in kelvins when focusing 850 spots of sunlight on a single 2" x 2" surface, we need to know the total