Problem 87 - Master Resource Book in JEE Main Physics

Question

An ice box used for keeping eatable cold has a total wall area of \(1 \mathrm{~m}^{2}\) and a wall thickness of \(5.0 \mathrm{~cm}\). The thermal conductivity of the ice box is \(K=0.01\) joule/metre- \({ }^{\circ} \mathrm{C}\). It is filled with ice at \(0^{\circ} \mathrm{C}\) along with eatables on a day when the temperature is \(30^{\circ} \mathrm{C}\). The latent heat of fusion of ice is \(334 \times 10^{3}\) joule \(/ \mathrm{kg}\). The amount of ice melted in one day is \((1\) day \(=86.400 \mathrm{~s})\) (a) \(776 \mathrm{~g}\) (b) \(7760 \mathrm{~g}\) (c) \(11520 \mathrm{~g}\) (d) \(1552 \mathrm{~g}\)

Step-by-Step Solution

Answer

Option (d) 1552 g.

Step 1: Identify Given Quantities

We have the following given values from the problem:- Total wall area of the ice box, \( A = 1 \text{ m}^2 \).- Wall thickness, \( d = 5.0 \text{ cm} = 0.05 \text{ m} \).- Thermal conductivity, \( K = 0.01 \text{ J/m-°C} \).- Temperature difference, \( \Delta T = 30^{\circ}C - 0^{\circ}C = 30^{\circ}C \).- Latent heat of fusion, \( L = 334 \times 10^3 \text{ J/kg} \).- Time duration for one day, \( t = 86400 \text{ s} \).

Step 2: Calculate Heat Transfer Through the Walls

The formula for heat transfer through a material is given by \[ Q = \frac{KA\Delta T t}{d} \]Substitute the known values:\[ Q = \frac{0.01 \times 1 \times 30 \times 86400}{0.05} \]Compute \( Q \):\[ Q = \frac{0.01 \times 30 \times 86400}{0.05} = 51840 \text{ J} \]

Step 3: Relate Melted Ice Mass to Heat Transfer

The energy required to melt ice is calculated using the equation\[ Q = mL \]where \( m \) is the mass of ice melted. Rearrange for \( m \):\[ m = \frac{Q}{L} \]Substitute the values for \( Q \) and \( L \):\[ m = \frac{51840}{334 \times 10^3} \approx 0.1552 \text{ kg} \]

Step 4: Convert Mass of Melted Ice to Grams

Convert the mass from kilograms to grams:\[ m = 0.1552 \text{ kg} \times 1000 \text{ g/kg} = 155.2 \text{ g} \]

Step 5: Verify Calculation and Choose the Closest Option

From the option provided, we see that the closest match to our calculated melted ice mass is \( 1552 \text{ g} \).

Key Concepts

Heat Transfer Equation

The heat transfer equation is key to understanding how heat moves through materials. In physics, it is represented by the formula:\[ Q = \frac{KA\Delta T t}{d} \]Where:

• \(Q\) is the heat transferred through the material, measured in joules (J).
• \(K\) is the thermal conductivity of the material. This value tells us how well the material can conduct heat. Higher values mean better heat conduction.
• \(A\) is the surface area through which the heat is transferred, given in square meters (m²).
• \(\Delta T\) is the temperature difference across the material in degrees Celsius (°C).
• \(t\) is the time for which heat transfer occurs, measured in seconds (s).
• \(d\) is the thickness of the material, given in meters (m).

This equation helps calculate the amount of energy transferred across a certain material over a period of time. In our ice box scenario, knowing this helps determine how much ice will melt because of the heat coming in from the exterior environment. It shows that thicker walls and lower conductivity materials slow down heat transfer, assisting in keeping the contents cooler for longer.

Phase Change and Latent Heat

When a substance changes its phase, such as from solid ice to liquid water, it requires or releases energy, depending on the direction of the change. This energy is called latent heat. The term \'latent\' refers to the hidden nature of this energy—it goes into effect without changing the temperature of the substance.In particular, the latent heat of fusion is relevant when discussing the melting of ice. It is the heat required to change a unit mass of a solid into a liquid without a temperature shift. For water/ice, it is quantified as:- Latent heat of fusion, \( L = 334 \times 10^3 \text{ J/kg} \)To find out how much ice will melt due to heat transfer, use the formula:\[ Q = mL \]Rearrange it to find the mass of ice:\[ m = \frac{Q}{L} \]Where:

• \(m\) is the mass of ice melted, in kilograms (kg).
• \(L\) is the latent heat of fusion, in joules per kilogram (J/kg).

This concept is crucial for calculating how much of a given substance will undergo a phase change when subjected to a particular amount of thermal energy, like in our exercise.

Units and Conversions in Physics

Understanding units and being able to convert them is crucial in solving physical problems. Physics often involves diverse units, and you need to ensure all units are compatible when plugging them into formulas. For instance, in the heat transfer equation and latent heat calculations, ensuring consistency in units is important:

• Area: Usually measured in square meters (m²).
• Thickness: Converts from centimeters to meters by dividing by 100.
• Time: Shown in seconds (s), as typical in scientific computations.
• Thermal Energy (Heat): Presented in joules (J).
• Mass: Usually given in kilograms (kg), but may need conversion to grams (g) for finer measurements, multiplying kg value by 1000.

Such conversions ensure precision in calculations. In the solved exercise, knowing the specific transformations was necessary, like converting the wall thickness from centimeters to meters or adjusting the mass of melted ice from kilograms to grams. Mastering these conversions can make handling physics problems more intuitive and easier to work through.