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Step-by-Step Solution
Step 1: Identify the Given Quantities
• Energy flux of light, $I = 25\,\text{W/cm}^2$ (this is the power per unit area).
• Area of the surface, $A = 25\,\text{cm}^2$.
• Time interval, $t = 40\,\text{min} = 40 \times 60 = 2400\,\text{s}$.
• Speed of light, $c = 3 \times 10^8\,\text{m/s}.$
Step 2: Calculate the Total Energy Incident on the Surface
The power (energy per unit time) incident on the surface is:
$P_{\text{incident}} = I \times A = 25 \,\text{W/cm}^2 \times 25 \,\text{cm}^2.$
This equals
$P_{\text{incident}} = 625\,\text{W} = 625\,\text{J/s}.$
Now, the total energy incident over 2400 s is:
$\Delta E = P_{\text{incident}} \times t = 625\,\text{J/s} \times 2400\,\text{s} = 1.5 \times 10^6\,\text{J}.$
Step 3: Relate Energy to Momentum for Completely Absorbed Light
For light that is completely absorbed by a surface, the momentum transferred $p$ is given by:
$p = \frac{\Delta E}{c}.$
Step 4: Compute the Momentum Transferred
Substitute $\Delta E = 1.5 \times 10^6\,\text{J}$ and $c = 3 \times 10^8\,\text{m/s}$:
$p = \frac{1.5 \times 10^6}{3 \times 10^8}\,\text{Ns} = 5.0 \times 10^{-3}\,\text{Ns}.$
Step 5: State the Final Answer
Therefore, the momentum transferred to the surface in 40 minutes is
$\boxed{5.0 \times 10^{-3}\,\text{Ns}.}$
