Question

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

• A. \(4 \, \text{cal/cm}^2 \text{ min}\)
• B. \(40 \, \text{cal/cm}^2 \text{ min}\)
• C. \(4 \, \text{J/m}^2 \text{ min}\)
• D. \(400 \, \text{cal/cm}^2 \text{ min}\)

Hint:

Always use latent heat for phase change problems: \(Q = mL\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The total heat required to melt the ice is calculated using the latent heat of fusion. The intensity (heat per unit area per unit time) is this total heat divided by the area of the lens and the time duration.
: Key Formula or Approach:
1. Heat required \( Q = m \cdot L_f \). (Latent heat of ice \( L_f = 80 \text{ cal/g} \))
2. Area of lens \( A = \pi r^2 = \pi (d/2)^2 \).
3. Intensity \( I = \frac{Q}{A \cdot t} \).
Step 2: Detailed Explanation:
- Heat required to melt 10 g of ice:
\[ Q = 10 \text{ g} \times 80 \text{ cal/g} = 800 \text{ cal} \]
- Time taken \( t = 10 \text{ min} \).
- Heat received per minute \( = 800 / 10 = 80 \text{ cal/min} \).
- Radius of the lens \( r = 5 / 2 = 2.5 \text{ cm} \).
- Area of the lens \( A = \pi \times (2.5)^2 \approx 3.14 \times 6.25 \approx 19.625 \text{ cm}^2 \).
- Heat per unit area per minute:
\[ \text{Value} = \frac{80 \text{ cal/min}}{19.625 \text{ cm}^2} \approx 4.07 \approx 4 \text{ cal/cm}^2 \text{ min} \]
Step 3: Final Answer:
The amount of heat received is \( 4 \text{ cal/cm}^2 \text{ min} \).