Question

Using Wien’s Law, estimate the wavelength at which these common materials radiate their peak brightness (show your work):

walls at 300 K:

ice water at 273 K :

hot water at 373 K :

skin at 310 K:

An incandescent light bulb at 3000 K

Answer 1

Using Wien's Law and the formula (Amax T = 2.898 × 10^-3 m·K), we calculated the peak emission wavelengths for walls at 9.66 μm, ice water at 10.62 μm, hot water at 7.77 μm, skin at 9.35 μm, and an incandescent light bulb at 0.966 μm.

Step-by-step explanation:

To estimate the peak wavelength at which different materials radiate their peak brightness using Wien's Law, we use the formula Amax T = 2.898 × 10^-3 m·K. Wien's Law states that the wavelength at which an object's emission is brightest is inversely proportional to its temperature in Kelvins.

Steps for calculating the peak emission wavelength:

1. Start by plugging in the given temperature (T) into Wien's Law formula.
2. Rearrange the formula to solve for the peak wavelength λmax (λmax = constant / T).
3. Perform the calculation to find λmax in meters, then convert to micrometers (μm) by multiplying by 10^6.

Let's calculate the peak emission wavelengths for the given materials:

• Walls at 300 K: λmax = 2.898 × 10^-3 m·K / 300 K = 9.66 × 10^-6 m or 9.66 μm
• Ice water at 273 K: λmax = 2.898 × 10^-3 m·K / 273 K = 10.62 × 10^-6 m or 10.62 μm
• Hot water at 373 K: λmax = 2.898 × 10^-3 m·K / 373 K = 7.77 × 10^-6 m or 7.77 μm
• Skin at 310 K: λmax = 2.898 × 10^-3 m·K / 310 K = 9.35 × 10^-6 m or 9.35 μm
• An incandescent light bulb at 3000 K: λmax = 2.898 × 10^-3 m·K / 3000 K = 0.966 × 10^-6 m or 0.966 μm

Answer 2

Step-by-step explanation:

Wien’s Law:

`\lambda _(max)\propto (1)/(temperature)`

The law states that the wavelength at which black body emits maximum heat is inversely proportional to the temperature of the body in Kelvins.

`\lambda _(max)=(2.9* 10^(-3))/(T)`

1) Walls at 300 K

`\lambda _(max)=(2.9* 10^(-3) mK)/(300 K)`

`\lambda _(max)=9.666* 10^(-6) m`

2) Ice water at 273 K :

`\lambda _(max)=(2.9* 10^(-3) mK)/(273 K)`

`\lambda _(max)=1.062* 10^(-5) m`

3) Hot water at 373 K :

`\lambda _(max)=(2.9* 10^(-3) mK)/(373 K)`

`\lambda _(max)=7.775* 10^(-6) m`

4)Skin at 310 K:

`\lambda _(max)=(2.9* 10^(-3) mK)/(310 K)`

`\lambda _(max)=9.355* 10^(-6) m`

5)An incandescent light bulb at 3000 K

`\lambda _(max)=(2.9* 10^(-3) mK)/(3000 K)`

`\lambda _(max)=9.666* 10^(-7) m`