Problem 52 - Arihant AIEEE Physics

Question

It takes 20 minutes to melt \(10 \mathrm{~g}\) of ice, when rays from the 'sun are focussed by a lens of diameter \(5 \mathrm{~cm}\) on to a block of ice. The heat received from the sun on \(1 \mathrm{~cm}^{2}\) per minute is : (Given: \(L=80 \mathrm{k} \mathrm{cal} / \mathrm{kg}\) ) (a) \(R=2.04 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}\) (b) \(R=3.04 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}\) (c) \(R=0.204 \mathrm{cal} / \mathrm{cm}^{2}-\min\) (d) \(R=204 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}\)

Step-by-Step Solution

Answer

(a) \( R = 2.04 \text{ cal/cm}^2\text{-min} \).

Step 1: Understanding the given data

We are given that it takes 20 minutes to melt 10 grams of ice. The latent heat of ice, \(L\), is 80 kcal/kg. We need to find the rate \(R\) of heat received per square centimeter per minute.

Step 2: Convert units for latent heat

The latent heat \(L\) is given in kcal/kg. Convert it to cal/g: \[L = 80 \text{kcal/kg} = 80,000 \text{cal/kg} = 80 \text{cal/g}\]

Step 3: Calculate total heat needed to melt the ice

The total heat \(Q\) needed to melt 10 grams of ice can be calculated using the formula: \[Q = m \times L = 10 \text{g} \times 80 \text{cal/g} = 800 \text{cal}\]

Step 4: Calculate the total area of the lens

The diameter of the lens is 5 cm, so the radius \(r\) is 2.5 cm. The area \(A\) is thus: \[A = \pi r^2 = \pi (2.5)^2 = 6.25\pi \approx 19.63 \text{cm}^2\]

Step 5: Calculate total energy received per minute

Since it takes 20 minutes to melt the ice: \[\text{Total energy received per minute} = \frac{Q}{20} = \frac{800 \text{cal}}{20} = 40 \text{cal/min}\]

Step 6: Calculate heat received per square centimeter per minute

The heat per square centimeter \(R\) is: \[R = \frac{\text{Total energy per minute}}{A} = \frac{40 \text{cal/min}}{19.63 \text{cm}^2} \approx 2.04 \text{cal/cm}^2\text{-min}\]

Step 7: Choose the correct option

Comparing the calculated value of \(R\) with the given options, the correct answer is: (a) \( R = 2.04 \text{ cal/cm}^2\text{-min} \).

Key Concepts

Latent Heat

In physics, latent heat refers to the heat energy required to change the state of a substance without changing its temperature. For example, when ice melts, it absorbs a certain amount of heat from its surroundings without any increase in temperature. This absorbed energy breaks the bonds between the water molecules in the ice, resulting in a phase change from solid to liquid.
In this exercise, we focus on the conversion of ice to water, a process defined by its latent heat of fusion. The latent heat of fusion for ice is given as 80 kcal/kg, which is equivalent to 80 cal/g. This value is crucial because it tells us the specific amount of energy needed to melt each gram of ice. Understanding how latent heat works is essential in solving problems related to heat transfer and phase changes, as it directly impacts the energy calculations required for the process.
By knowing the mass of the ice and the latent heat, you can compute the total energy needed to achieve the phase change, as done in the solution provided.

Thermodynamics

Thermodynamics is the branch of physics that studies heat, work, and the forms of energy involved in physical and chemical processes. It is deeply rooted in concepts such as temperature, energy conservation, and the laws dictating the behavior of heat in different systems.
The problem we are dealing with embodies a practical application of thermodynamics, where principles such as energy conservation play a pivotal role. The energy from the sun is focused through a lens, leading to the melting of the ice. Here, it's important to note that the energy received must be equal to the energy required to melt the ice, as dictated by the first law of thermodynamics.
In this scenario, the heat being transferred from the sunlight to the ice is a perfect illustration of thermal energy conversion. As we calculate how much energy is needed to phase change the solid ice to liquid, we're essentially applying fundamental thermodynamic principles to solve the problem.

Physics Problem Solving

Solving physics problems, especially in thermodynamics, requires a systematic approach to break down the given information and apply relevant formulas. This exercise is a great example of how to efficiently solve a practical physics problem.
Here are the key steps in physics problem-solving for such scenarios:

• Understand the problem statement and what is being asked. Identify knowns and unknowns.
• Convert units to simplify calculations. Consistency in units is crucial to obtaining the correct results.
• Apply the relevant formulae, keeping in mind the physical principles that justify their use, like latent heat in this case.
• Perform step-by-step calculations, verifying each solution stage for accuracy.
• Compare the results with the given options or expected results to ensure correctness.

Being methodical and step-by-step ensures that no detail is overlooked, leading to the correct resolution of complex problems like the one mentioned in the exercise.