Problem 71 - Master Resource Book in JEE Main Physics

Question

The only possibility of heat flow in a themros flask is through its cork which is \(75 \mathrm{~cm}^{2}\) in area and \(5 \mathrm{~cm}\) thick its thermal conductivity is \(0.075 \mathrm{cal} / \mathrm{cm} \sec ^{\circ} \mathrm{C}\). The outside temperatue is \(40^{\circ} \mathrm{C}\) and latent heat of ice is \(80 \mathrm{cal} \mathrm{g}^{-1}\). Time taken by \(500 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) in the flask to melt into water at \(0^{\circ} \mathrm{C}\) is (a) \(2.47 \mathrm{~h}\) (b) \(4.27 \mathrm{~h}\) (c) \(7.42 \mathrm{~h}\) (d) \(4.82 \mathrm{~h}\)

Step-by-Step Solution

Answer

Reassess calculated steps, as 0.247 hours doesn't match given options.

Step 1: Understanding the Concept of Heat Transfer

The heat transfer through the flask's cork is due to thermal conductivity. The heat by conduction is given by the formula \( Q = \frac{k \cdot A \cdot (T_1 - T_2) \cdot t}{d} \), where \( Q \) is the heat transfer in calories, \( k \) is the thermal conductivity, \( A \) is the area, \( (T_1 - T_2) \) is the temperature difference, \( t \) is the time, and \( d \) is the thickness of the material.

Step 2: Express Heat Required to Melt Ice

The heat required to melt 500 g of ice is given by the formula \( Q = mL \), where \( Q \) is the heat in calories, \( m \) is the mass of the ice, and \( L \) is the latent heat of fusion of ice. Here, \( m = 500 \text{ g} \) and \( L = 80 \text{ cal/g} \). So, \( Q = 500 \times 80 = 40000 \text{ cal} \).

Step 3: Calculate Heat Flow Rate Through the Cork

Substitute the given values into the formula for heat transfer to find the rate of heat flow. Given \( k = 0.075 \text{ cal/cm sec }^\circ\text{C} \), \( A = 75 \text{ cm}^2 \), \( T_1 = 40^\circ\text{C} \), \( T_2 = 0^\circ\text{C} \), and \( d = 5 \text{ cm} \):\[ \frac{Q}{t} = \frac{0.075 \cdot 75 \cdot (40 - 0)}{5} = 45 \text{ cal/sec} \].

Step 4: Calculate Time for Ice to Melt

Use the relation \( Q = \frac{Q}{t} \cdot t \) to find \( t \). Rearranging gives \( t = \frac{Q}{\frac{Q}{t}} = \frac{40000}{45} \). Thus, \( t \approx 888.89 \text{ seconds}\). Convert seconds to hours: \( t = \frac{888.89}{3600} \approx 0.247 \text{ hours} \). Since this result doesn't match any provided options, let's go back and double-check the arithmetics and conversions for potential errors.

Step 5: Re-evaluate Calculation and Resolve

After reassessing and ensuring consistent calculations and units, if repeated verification upholds our steps and math, logical next step involves possible oversight on conversion from seconds to hours, incorrect input values or internal error in given choices.

Key Concepts

Thermal Conductivity

Thermal conductivity is a property that measures a material's ability to conduct heat. It is denoted by the symbol \( k \). In simpler terms, it tells us how quickly heat can travel through a material. For example, metals usually have high thermal conductivity and can transfer heat rapidly, whereas materials like cork or rubber have low thermal conductivity, meaning they slow down the transfer of heat.
In the exercise, the cork's thermal conductivity is given as \( 0.075 \text{ cal/cm sec }^{\circ}\text{C} \). This low value is typical because cork is a good thermal insulator. It means that the cork doesn't let heat pass through easily. This property helps keep the contents of a thermos flask at a steady temperature, whether it's keeping your ice solid or your coffee hot.

• Area \( A \): The effectiveness of thermal conductivity also depends on the cross-sectional area through which the heat is passing. A larger area can transfer more heat.
• Thickness \( d \): The thicker the material, the harder it is for heat to flow through, making it a crucial factor in insulating properties.

Latent Heat of Fusion

Latent heat of fusion is a physical property that explains the amount of heat required to change a solid into a liquid without changing its temperature. For ice, this value is \( 80 \text{ cal/g} \). This means that to melt one gram of ice into water, we need to supply 80 calories of heat. During this process, the temperature of the ice remains constant at \( 0^{\circ}\text{C} \).
In the problem, you have \( 500 \text{ g} \) of ice, and the latent heat of fusion is a critical factor because it tells us exactly how much heat energy is required. You would need \( 80 \text{ cal/g} \times 500 \text{ g} = 40000 \text{ cal} \) to melt the ice completely into water at the same temperature. This energy doesn't increase the temperature but changes the state from solid to liquid.

• No temperature change: The energy absorbed is used entirely for the phase transformation rather than changing the temperature.
• Ice to water: This process of melting is crucial in various natural and industrial contexts, illustrating how thermal energy manages state changes.

Heat Transfer Equations

Heat transfer is a vital concept in physics that helps us understand how energy moves from one part of a system to another. The amount of heat transferred over time is found using specific equations. The main equation in this context is related to thermal conduction, which in the exercise is given by:\[ Q = \frac{k \cdot A \cdot (T_1 - T_2) \cdot t}{d} \]In this formula, \( Q \) represents the amount of heat transferred, \( k \) is the thermal conductivity, \( A \) is the area through which heat is being conducted, \( T_1 - T_2 \) is the temperature difference across the material, \( t \) is the time, and \( d \) is the material's thickness.
This formula indicates that the heat transfer rate depends on multiple factors, including temperature gradient and material properties. The rate of heat flow is essential to finding out how long a process, like melting ice, will take under certain conditions. In this exercise, once the rate \( \frac{Q}{t} = 45 \text{ cal/sec} \) is found, we can use it with the known heat \( Q \) to compute the time required for the ice to melt:

• Time \( t \): Calculated by rearranging the equation to \( t = \frac{Q}{\frac{Q}{t}} \)
• Consistency: Ensuring all units match and calculations follow correct arithmetic is crucial.

These equations form the basis for solving many practical problems involving heat management and thermal systems.