Problem 54 - Master Resource Book in JEE Main Physics

Question

Water of volume \(2 \mathrm{~L}\) in a container is heated with a coil of \(1 \mathrm{~kW}\) at \(27^{\circ} \mathrm{C}\). The lid of the container is open and energy dissipates at rate of \(160 \mathrm{~J} / \mathrm{s}\). In how much time temperature will rise from \(27^{\circ} \mathrm{C}\) to \(77^{\circ} \mathrm{C}\) ? [Given specific heat of water is \(4.2 \mathrm{~kJ} / \mathrm{kg}\) ] (a) \(8 \min 20 \mathrm{~s}\) (b) \(6 \min 2 \mathrm{~s}\) (c) 7 min (d) \(14 \mathrm{~min}\)

Step-by-Step Solution

Answer

(a) 8 min 20 s

Step 1: Determine Heat Required for Temperature Change

The heat required to increase the temperature of the water can be calculated using the formula \( Q = mc\Delta T \), where \( m \) is the mass of the water, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Given:- Volume of water \( V = 2 \text{ L} = 2 \text{ kg} \) (since the density of water is \( 1 \text{ kg/L} \))- \( c = 4.2 \text{ kJ/kg°C} = 4200 \text{ J/kg°C} \)- \( \Delta T = 77°C - 27°C = 50°C \)Therefore, \[ Q = 2 \times 4200 \times 50 = 420,000 \text{ J} \]

Step 2: Calculate Net Power Input to the System

The net power input is the power supplied by the coil minus the power lost.Given:- Power of coil, \( P_{ ext{coil}} = 1000 \text{ W} = 1000 \text{ J/s} \)- Energy dissipation rate, \( P_{ ext{loss}} = 160 \text{ J/s} \)Net power input:\[ P_{ ext{net}} = P_{ ext{coil}} - P_{ ext{loss}} = 1000 - 160 = 840 \text{ J/s} \]

Step 3: Calculate Time Required to Heat the Water

To find the time required, use the relation \( Q = P \times t \), where \( Q \) is the heat energy required, \( P \) is the net power input, and \( t \) is the time in seconds.We have:\[ 420,000 \text{ J} = 840 \text{ J/s} \times t \]Solving for \( t \):\[ t = \frac{420,000}{840} = 500 \text{ s} \]

Step 4: Convert Time from Seconds to Minutes and Seconds

The calculated time in seconds needs to be converted to minutes and seconds.We know:- 1 minute = 60 secondsThus,\[ 500 \text{ s} = 8 \text{ minutes} + 20 \text{ seconds} \]

Step 5: Final Step: Verify the Solution

Ensure that all calculations align with the given problem constraints and that the converted time aligns with one of the provided choices. We have calculated the time to be 8 minutes and 20 seconds, which matches option (a), confirming the solution.

Key Concepts

Heat Transfer

Heat transfer is the process by which thermal energy moves from a hotter object to a cooler one. In this exercise, heat is transferred from a coil to water, increasing the water's temperature. The coil, acting as a heater, provides the energy needed to raise the temperature of the water. This process involves various mechanisms such as conduction, convection, and radiation. In our scenario, the focus is on the energy supplied by the coil and dissipated to the surroundings. The net heat provided accounts for the energy lost to the environment, an important factor in determining how efficiently the water is heated.

• Conduction: Movement of heat through a material without physical movement of the substance itself.
• Convection: Transfer of heat through a fluid (liquid or gas) as fluid moves from one place to another.
• Radiation: Transfer of energy through electromagnetic waves. No medium is required.

Understanding these mechanisms helps explain how heat from the coil warms the water despite continuous energy loss to the environment.

Specific Heat Capacity

Specific heat capacity is a key concept in understanding heat transfer as it quantifies the amount of heat needed to raise the temperature of a unit mass of a substance by one degree Celsius. For water, this value is given as 4.2 kJ/kg°C in our problem. It tells us that water requires 4.2 kJ of energy to increase the temperature of one kilogram by 1°C. This property of water indicates its ability to absorb heat without a large temperature change, making it useful for heating systems.
In our task, the specific heat capacity formula is:\[ c = \frac{Q}{m \times \Delta T} \]Where:

• \( Q \) is the total amount of heat absorbed or released by the system.
• \( m \) is the mass of the water.
• \( \Delta T \) is the change in temperature.

The amount of heat required to raise the water temperature can be calculated using this formula, helping predict how much energy the coil must supply.

Power and Energy Calculations

Power measures how quickly energy is used or transmitted in a system. It is calculated as energy divided by time and is measured in watts (W), where 1 W = 1 J/s. In the exercise, the coil's power output is 1 kW, and the specific task is to find how long it takes to heat the water, given continuous energy loss.
The net power input into the system is the difference between the power output from the coil and the energy dissipation rate:\[ P_{ \text{net} } = P_{ \text{coil} } - P_{ \text{loss } } \]With the net power, we use the relation:\[ Q = P \times t \]To solve for time \( t \):\[ t = \frac{Q}{P_{ \text{net} }} \]This provides the duration required for the desired temperature increase given the conditions of heat input and loss. The end calculation involves converting the time from seconds to minutes and determining if it aligns with the provided options. Understanding power and energy dynamics is essential for efficient heat management in systems similar to the exercise's dynamic.