The rays of the sun are focused on a piece of ice through a lens of diameter 5 cm , as a result of which 10g ice melts in 10 minutes. The amount of heat received from the sun per unit area per minute is

Options

A. `4 "cal cm"^-2min.^(-1)`
B. `40 "cal cm"^-2min.^(-1)`
C. `4 J cm ^-2min.^(-1)`
D. `400 "cal cm "^-2min.^(-1)`

Answer

To solve the problem, we need to find the amount of heat received from the sun per unit area per minute, which is referred to as intensity (I).

### Step-by-Step Solution:

1. **Identify the given data:**
- Diameter of the lens (D) = 5 cm
- Mass of ice melted (m) = 10 g
- Time (t) = 10 minutes

2. **Calculate the radius of the lens:**
\[ \text{Radius} (r) = \frac{D}{2} = \frac{5 \text{ cm}}{2} = 2.5 \text{ cm} \]

3. **Calculate the area (A) of the lens:**
The area of a circle is given by the formula:
\[ A = \pi r^2 \]
Substituting the radius:
\[ A = \pi (2.5 \text{ cm})^2 = \pi (6.25 \text{ cm}^2) = 6.25\pi \text{ cm}^2 \]

4. **Calculate the heat required to melt the ice:**
The latent heat of fusion of ice is given as 80 calories per gram. Therefore, the total heat (Q) required to melt 10 g of ice is:
\[ Q = m \times L = 10 \text{ g} \times 80 \text{ cal/g} = 800 \text{ cal} \]

5. **Relate heat to intensity:**
The intensity (I) can be defined as:
\[ I = \frac{Q}{A \cdot t} \]
Here, we need to convert the time from minutes to seconds for standard units, but since we want the answer in per minute, we can keep it as is:
\[ I = \frac{800 \text{ cal}}{6.25\pi \text{ cm}^2 \cdot 10 \text{ min}} \]

6. **Calculate the intensity:**
\[ I = \frac{800}{62.5\pi} \text{ cal/cm}^2/\text{min} \]
Simplifying this:
\[ I = \frac{800}{62.5\pi} = \frac{800}{62.5 \times 3.14} \approx \frac{800}{196.25} \approx 4.08 \text{ cal/cm}^2/\text{min} \]

### Final Answer:
The amount of heat received from the sun per unit area per minute is approximately **4.08 cal/cm²/min**.