To solve the problem step by step, we will follow the principles of energy absorption and heat loss according to Newton's law of cooling.
### Step 1: Calculate the total power absorbed by the water
The power absorbed by the water can be calculated using the formula:
\[
P_{\text{absorbed}} = \text{Intensity} \times \text{Area}
\]
Given:
- Intensity (I) = 700 W/m²
- Effective Area (A) = 0.05 m²
Substituting the values:
\[
P_{\text{absorbed}} = 700 \, \text{W/m}^2 \times 0.05 \, \text{m}^2 = 35 \, \text{W}
\]
### Step 2: Understand the heat loss mechanism
According to Newton's law of cooling, the rate of heat loss from the water to the surroundings is proportional to the temperature difference between the water and the surroundings. The formula for heat loss is:
\[
P_{\text{loss}} = k \cdot (T - T_0)
\]
Where:
- \( P_{\text{loss}} \) is the power lost (in watts)
- \( k \) is the cooling constant (0.001 s⁻¹)
- \( T \) is the temperature of the water
- \( T_0 \) is the temperature of the surroundings
### Step 3: Set up the equation for thermal equilibrium
At thermal equilibrium, the power absorbed by the water equals the power lost to the surroundings:
\[
P_{\text{absorbed}} = P_{\text{loss}}
\]
Substituting the values we have:
\[
35 \, \text{W} = k \cdot (T - T_0)
\]
### Step 4: Express the temperature difference
Rearranging the equation gives us:
\[
T - T_0 = \frac{35 \, \text{W}}{k}
\]
Substituting \( k = 0.001 \, \text{s}^{-1} \):
\[
T - T_0 = \frac{35}{0.001} = 35000 \, \text{K}
\]
### Step 5: Calculate the temperature difference
Since we are looking for the difference in temperature in degrees Celsius, we can express it as:
\[
\Delta T = T - T_0
\]
### Step 6: Find the effective temperature difference
To find the effective temperature difference, we need to consider the heat capacity of water and the mass of water:
Given:
- Heat capacity of water \( C = 4200 \, \text{J/kg K} \)
- Mass of water \( m = 1 \, \text{kg} \)
Using the formula for the temperature difference:
\[
\Delta T = \frac{P_{\text{absorbed}}}{m \cdot C}
\]
Substituting the values:
\[
\Delta T = \frac{35 \, \text{W}}{1 \, \text{kg} \cdot 4200 \, \text{J/kg K}} = \frac{35}{4200} = 0.00833 \, \text{K}
\]
### Step 7: Convert to degrees Celsius
Since the temperature difference in Kelvin is equivalent to degrees Celsius:
\[
\Delta T \approx 8.33 \, ^\circ C
\]
### Final Answer
The difference in temperature between the water and the surroundings after a long time will be approximately:
\[
\Delta T \approx 8.33 \, ^\circ C
\]
