How Long to Heat Water with Blackbody Radiation?

[moogull]

Homework Statement

Raising the temperature of 1$cm^{3}$ of water (1 gram, water has a heat cap. of 4.2 J/g*K) using energy from a cavity filled with black body radiation. The water is to change from 299K to 300K. The radiation is initially at 450K.
If the cavity has a volume of 0.1 $m^{3}$ how long would you need to have the heating coils turned on to heat the water? (voltage of coils is 240V, and current is 20A)

Homework Equations

P=IV
Energy of a photon gas = ($\frac{8\pi^{5}*k^4}{15c^3*h^3}$)V$T^{4}$
radiative flux = $\sigma$$T^{4}$ sigma is stefan-boltzmann constant

The Attempt at a Solution

I want to say that the energy input from the coils should be enough and the time would be 4.2J /480V*20A, but that can't be it. Honestly I'm totally lost on this one because the radiative flux would require that you know a surface area, but one is not given (and maybe that the volume doesn't even matter), any thoughts?

 

[mark726]

Hello, thank you for your post. To calculate the time needed to heat the water, we can use the equation for energy transfer through radiation:

Q = \sigma A (T2^4 - T1^4)t

Where Q is the energy transferred, \sigma is the Stefan-Boltzmann constant, A is the surface area of the cavity, T2 is the final temperature (in Kelvin), T1 is the initial temperature (in Kelvin), and t is the time (in seconds).

Since we are given the volume of the cavity (0.1 m^3), we can assume that it is a cube with each side measuring 0.464 m (V = l*w*h, so h = V/(l*w)). This means that the surface area of the cavity can be calculated as:

A = 6 * (0.464 m)^2 = 1.08 m^2

Now, we can plug in the values for the temperatures and the surface area into the equation:

Q = (5.67*10^-8 W/m^2*K^4) * (1.08 m^2) * (300^4 - 299^4) * t

Solving for t, we get:

t = 4.2 J / (480V * 20A) = 0.00175 seconds

Therefore, it would take approximately 0.00175 seconds for the heating coils to raise the temperature of the water from 299K to 300K.