Earth receives `1400 W//m^2` of solar power. If all the solar energy falling on a lens of area `0.2m^2` is focused on to a block of ice of mass 280 grams, the time taken to melt the ice will be….. Minutes. (Latent heat of fusion of ice=`3.3xx10^5J//kg)`

Options

A. 300s
B. 330s
C. 200s
D. 150s

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the energy falling on the lens
The solar power received by the Earth is given as \( 1400 \, \text{W/m}^2 \). The area of the lens is \( 0.2 \, \text{m}^2 \).

The total power \( P \) falling on the lens can be calculated using the formula:
\[ P = \text{Solar Power} \times \text{Area} \]
\[ P = 1400 \, \text{W/m}^2 \times 0.2 \, \text{m}^2 = 280 \, \text{W} \]

### Step 2: Convert the mass of ice to kilograms
The mass of the block of ice is given as \( 280 \, \text{g} \). We need to convert this to kilograms:
\[ m = \frac{280 \, \text{g}}{1000} = 0.28 \, \text{kg} \]

### Step 3: Calculate the heat required to melt the ice
The latent heat of fusion of ice is given as \( 3.3 \times 10^5 \, \text{J/kg} \). The heat \( Q \) required to melt the ice can be calculated using the formula:
\[ Q = m \times L \]
where \( L \) is the latent heat of fusion.

Substituting the values:
\[ Q = 0.28 \, \text{kg} \times 3.3 \times 10^5 \, \text{J/kg} = 92400 \, \text{J} \]

### Step 4: Calculate the time taken to melt the ice
The power \( P \) is also defined as the energy per unit time:
\[ P = \frac{Q}{T} \]
Rearranging the formula to find time \( T \):
\[ T = \frac{Q}{P} \]
Substituting the values we calculated:
\[ T = \frac{92400 \, \text{J}}{280 \, \text{W}} = 330 \, \text{seconds} \]

### Step 5: Convert time from seconds to minutes
To convert seconds to minutes, we divide by 60:
\[ T = \frac{330 \, \text{seconds}}{60} = 5.5 \, \text{minutes} \]

### Final Answer
The time taken to melt the ice is **5.5 minutes**.

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