Earth receives 1400 W/m^2 of solar power. If all the solar energy falling on a lens of area 0.2 m^2

Q: Why do we need to convert the mass of ice from grams to kilograms?

The latent heat of fusion is given in $\text{J/kg}$. To ensure our units are consistent and the final energy is in Joules, the mass must also be in kilograms.

Q: What does the solar intensity of $1400 \text{ W/m}^2$ actually mean?

It means that every square meter of surface area exposed directly to the sun receives $1400 \text{ Joules}$ of energy every single second.

Q: Why do we multiply intensity by area to get power?

Intensity is power per unit area ($I = P/A$). Therefore, to find the total power collected by a specific area (like our lens), we multiply the intensity by that area.

Q: Is all the energy from the lens used to melt the ice?

In this idealized physics problem, yes. We assume $100\%$ efficiency, meaning no heat is lost to the surrounding air and all focused solar energy goes directly into melting the ice.

Q: Why don't we use specific heat capacity ($mc\Delta T$) here?

The problem assumes the ice is already at its melting point ($0^\circ\text{C}$) and only asks for the time to *melt* it. Melting is a phase change, which requires latent heat ($mL$), not specific heat, which is for changing temperature.

Question

Earth receives $1400	ext{ W/m}^2$ of solar power. If all the solar energy falling on a lens of area $0.2	ext{ m}^2$ is focussed onto a block of ice of mass $280	ext{ g}$, the time taken to melt the ice will be ...... minutes. (Latent heat of fusion of ice $= 3.3 	imes 10^5	ext{ J/kg}$)

Sigma Percentile · Accepted Answer

### The Power of the Sun

Imagine you are holding a magnifying glass on a bright, sunny day. The sun's rays, carrying immense energy, hit the lens and get focused onto a single, intense point. In our problem, that point is a block of ice.

Our ultimate goal is to find out exactly how long it takes for this focused solar energy to completely melt the ice block.

To solve this, we need to break the problem into two logical parts: finding the total energy required to melt the ice, and finding the rate at which the sun is supplying that energy.

### The Energy to Melt

First, let us figure out how much heat energy is actually needed to melt this block of ice.

We know from the principles of calorimetry that the heat required for a phase change, like melting, is given by the mass of the substance multiplied by its latent heat of fusion.

The master equation for this is:
$$Q = mL$$

Let us plug in the numbers. The mass of the ice is given as $280 	ext{ g}$. However, our latent heat is provided in $	ext{J/kg}$. To maintain consistent units, we must convert the mass to kilograms, which gives us $0.28 	ext{ kg}$.

We multiply this by the latent heat, $3.3 	imes 10^5 	ext{ J/kg}$:
$$Q = (0.28 	ext{ kg}) 	imes (3.3 	imes 10^5 	ext{ J/kg})$$

Now, let us do the math. Multiplying these values together, we find the total energy target we need to hit:
$$Q = 9.24 	imes 10^4 	ext{ J}$$

### Capturing Solar Power

Where is this energy coming from? It is coming directly from the sun! The lens acts as an energy collector.

The total power captured by the lens is the solar intensity, $I$, multiplied by the area of the lens, $A$. This tells us how much energy is arriving every single second.
$$P = I 	imes A$$

The problem states the solar intensity is $1400 	ext{ W/m}^2$, and the lens has an area of $0.2 	ext{ m}^2$. Let us substitute these values into our power equation:
$$P = (1400 	ext{ W/m}^2) 	imes (0.2 	ext{ m}^2)$$

Calculating this gives us $280 	ext{ W}$. Remember, a Watt is simply a Joule per second. 
$$P = 280 	ext{ J/s}$$

This means $280 	ext{ Joules}$ of energy are being pumped into the ice block every single second.

### Bringing It All Together

We now know the total energy required, $Q$, and we know the rate at which energy is supplied, $P$.

To find the total time taken, $t$, we simply divide the total energy by the power. It is exactly like knowing the total distance and your speed to find the time.
$$t = \frac{Q}{P}$$

Let us bring back the values we calculated. We substitute $9.24 	imes 10^4 	ext{ J}$ for $Q$, and $280 	ext{ J/s}$ for $P$:
$$t = \frac{9.24 	imes 10^4 	ext{ J}}{280 	ext{ J/s}}$$

Dividing these numbers gives us exactly $330$. Since our power was in Joules per second, this time is in seconds.
$$t = 330 	ext{ s}$$

The question specifically asks for the time in minutes. To convert seconds to minutes, we divide by $60$:
$$t = \frac{330}{60} 	ext{ minutes}$$

$$t = 5.5 	ext{ minutes}$$

And there we have it! By understanding how energy is transferred and used for phase changes, we have found that it takes exactly **$5.5 	ext{ minutes}$** to melt the ice block. A beautiful application of thermodynamics!

Enter Numerical Value:

Visualized Solution

\text{The Physical Setup}

\text{Heat Required to Melt Ice}

\text{Substituting Values for } Q

\text{Calculating } Q

\text{Power Received by the Lens}

\text{Substituting Values for } P

\text{Calculating } P

\text{Relating Energy, Power, and Time}

\text{Substituting Values for } t

\text{Calculating Time in Seconds}

\text{Converting Time to Minutes}

\text{Final Answer}

The Sigma Insight: Calorimetry

The Power of the Sun

The Energy to Melt

Capturing Solar Power

Bringing It All Together

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