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`.)

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Power Received by the Lens
The power received by the lens can be calculated using the formula:
\[ \text{Power} = \text{Solar Power} \times \text{Area} \]
Given:
- Solar Power = \(1400 \, \text{W/m}^2\)
- Area = \(0.2 \, \text{m}^2\)

Substituting the values:
\[ \text{Power} = 1400 \, \text{W/m}^2 \times 0.2 \, \text{m}^2 = 280 \, \text{W} \]

### Step 2: Convert the Mass of Ice to Kilograms
The mass of ice is given as \(280 \, \text{grams}\). We need to convert this to kilograms:
\[ \text{Mass} = 280 \, \text{g} = 0.280 \, \text{kg} \]

### Step 3: Calculate the Heat Required to Melt the Ice
The heat required to melt the ice can be calculated using the formula:
\[ Q = M \times L \]
Where:
- \(Q\) = Heat required
- \(M\) = Mass of ice = \(0.280 \, \text{kg}\)
- \(L\) = Latent heat of fusion of ice = \(3.3 \times 10^5 \, \text{J/kg}\)

Substituting the values:
\[ Q = 0.280 \, \text{kg} \times 3.3 \times 10^5 \, \text{J/kg} = 92400 \, \text{J} \]

### Step 4: Calculate the Time Taken to Melt the Ice
The time taken to melt the ice can be calculated using the formula:
\[ Q = P \times t \]
Where:
- \(Q\) = Heat required = \(92400 \, \text{J}\)
- \(P\) = Power = \(280 \, \text{W}\)
- \(t\) = Time in seconds

Rearranging the formula to find time:
\[ t = \frac{Q}{P} = \frac{92400 \, \text{J}}{280 \, \text{W}} = 330 \, \text{s} \]

### Step 5: Convert Time from Seconds to Minutes
To convert seconds into minutes:
\[ \text{Time in minutes} = \frac{330 \, \text{s}}{60} \approx 5.5 \, \text{minutes} \]

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

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