Problem 73 - Master Resource Book in JEE Main Physics

Question

A liquid is filled in a container which is kept in a room whose temperature is \(20^{\circ} \mathrm{C}\). When temperature of liquid is \(80^{\circ} \mathrm{C}\), it emits heat at the rate of \(45 \mathrm{cals}^{-1}\), When temperature of liquid falls to \(40^{\circ} \mathrm{C}\), its rate of heat loss will be (a) \(15 \mathrm{cals}^{-1}\) (b) \(30 \mathrm{cals}^{-1}\) (c) \(45 \mathrm{cal} \mathrm{s}^{-1}\) (d) \(60 \mathrm{cal}^{-1}\)

Step-by-Step Solution

Answer

15 cals s^{-1} (option a)

Step 1: Understand Newton's Law of Cooling

Newton's Law of Cooling states that the rate of heat loss of a body is proportional to the difference in its temperature and the ambient temperature. Mathematically, it can be expressed as \( \frac{dQ}{dt} = -k(T - T_{room}) \), where \( k \) is a proportionality constant, \( T \) is the temperature of the object, \( T_{room} \) is the room temperature, and \( \frac{dQ}{dt} \) is the rate of heat loss.

Step 2: Determine Proportionality Constant

From the problem, when the temperature of the liquid is \( 80^{\circ} \text{C} \), the rate of heat loss \( \frac{dQ}{dt} \) is \( 45 \text{ cals s}^{-1} \). Using the formula: \( \frac{dQ}{dt} = k(T - T_{room}) \), we have:\[k(80 - 20) = 45\]Simplify to find \( k \):\[k \times 60 = 45 \quad \Rightarrow \quad k = \frac{45}{60} = \frac{3}{4}\]

Step 3: Calculate Rate of Heat Loss at 40°C

Now, substitute the temperature \( 40^{\circ} \text{C} \) into the formula to find the rate of heat loss:\[\frac{dQ}{dt} = \frac{3}{4}(40 - 20)\]Simplify:\[\frac{dQ}{dt} = \frac{3}{4} \times 20 = 15 \text{ cals s}^{-1}\]

Step 4: Select the Correct Option

Based on our calculation, the rate of heat loss when the temperature drops to \( 40^{\circ} \text{C} \) is \( 15 \text{ cals s}^{-1} \). Therefore, the correct answer is option (a) \( 15 \text{ cals s}^{-1} \).

Key Concepts

Rate of Heat Loss

Understanding the rate of heat loss is crucial when studying Newton's Law of Cooling. This rate refers to how quickly a body loses heat to its surroundings. In physics, the rate of heat loss is typically represented as \( \frac{dQ}{dt} \), where \( Q \) is the heat energy and \( t \) is time. This formula helps us quantify how much heat is lost over time, allowing for effective thermal management and energy calculations in real-world applications.
This concept is particularly important in various scientific fields, such as thermodynamics and engineering. By calculating the rate of heat loss, engineers can design systems that are more energy-efficient, ensuring that they maintain their required temperatures for longer periods. Furthermore, it also assists in understanding the thermal behavior of substances subjected to temperature changes, like liquids in a heated container.

• Helps in quantifying heat exchange.
• Vital in energy efficiency designs.
• Applicable in thermodynamics and engineering calculations.

Proportionality Constant

The proportionality constant, denoted as \( k \) in Newton’s Law of Cooling, acts as a crucial factor in understanding how efficiently an object loses heat. This constant links the rate of heat loss directly to the temperature difference between the object and its surroundings. Essentially, it quantifies the relationship, allowing us to calculate specific heat loss rates for different conditions.
For instance, in the given problem, when the liquid is at \( 80^{\circ} \)C, substituting into the formula helps us solve for \( k \) by dividing the known rate of heat loss by the temperature difference. This constant remains specific to each material or condition, altering according to the environmental parameters and physical properties.

• Determines efficiency of heat loss.
• Unique to specific materials and conditions.
• Facilitates calculation of rate of heat loss.

Temperature Difference

The temperature difference between an object and its surroundings is pivotal in determining how fast the object will lose heat, according to Newton's Law of Cooling. This difference is the driving force behind the heat exchange process. The larger the temperature difference, the higher the rate of heat loss.
In practical terms, if you have a cup of coffee that's much hotter than room temperature, it will cool down faster than a cup that's only slightly warmer than room temperature. The formula \( T - T_{room} \) in Newton's Law helps us quantify this difference to understand its impact on the rate of heat loss.

• Primary factor influencing heat exchange rate.
• Larger differences imply faster cooling.
• Fundamental to many heat transfer calculations.