Problem 23 - Master Resource Book in JEE Main Physics

Question

A child running at a temperature of \(101^{\circ} \mathrm{F}\) is given an antipyrin (i.e., medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to \(98^{\circ} \mathrm{F}\) in \(20 \mathrm{~min}\), what is the average rate of extra evaporation caused by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of child is \(30 \mathrm{~kg}\). The specific heat of the human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about \(580 \mathrm{cal} / \mathrm{g}\). (a) \(4.31 \mathrm{~g} / \mathrm{min}\) (b) \(4.31 \mathrm{~g} / \mathrm{s}\) (c) \(2.31 \mathrm{~g} / \mathrm{min}\) (d) \(2.31 \mathrm{~g} / \mathrm{s}\)

Step-by-Step Solution

Answer

The average rate of extra evaporation caused by the drug is \(4.31 \text{ g/min}\), so the answer is (a).

Step 1: Calculate Heat Loss

The initial and final temperatures are given as \(101^{\circ} \mathrm{F}\) and \(98^{\circ} \mathrm{F}\) respectively. We need to first convert these temperatures to Celsius because the specific heat is given based on water's properties:\[T_{\text{initial}} = \frac{5}{9} \times (101 - 32) = 38.33^{\circ} \mathrm{C}\]\[T_{\text{final}} = \frac{5}{9} \times (98 - 32) = 36.67^{\circ} \mathrm{C}\]So, the temperature change \(\Delta T = 38.33 - 36.67 = 1.66^{\circ} \mathrm{C}\).The specific heat of the human body (same as that of water) is \(1 \text{ cal/g°C}\).Total heat loss, \(Q\), can be calculated as:\[Q = m \cdot c \cdot \Delta T = 30,000 \text{ g} \times 1 \text{ cal/g°C} \times 1.66 = 49,800 \text{ cal}\]

Step 2: Calculate Mass of Evaporated Water

To determine the mass of water evaporated, we use the latent heat of evaporation: \(\text{Latent heat of evaporation} = 580 \text{ cal/g} \)Using the heat loss calculated, the mass \(m_{\text{water}}\) evaporated is:\[m_{\text{water}} = \frac{Q}{L} = \frac{49,800 \text{ cal}}{580 \text{ cal/g}} = 85.86 \text{ g}\]

Step 3: Calculate Average Rate of Evaporation

The total mass of water evaporated \(85.86 \text{ g}\) in \(20\) minutes corresponds to an average evaporation rate. \[\text{Average rate} = \frac{85.86 \text{ g}}{20 \text{ min}} = 4.293 \text{ g/min}\]Rounding off, we find that:\[\text{Average rate} \approx 4.31 \text{ g/min}\]

Step 4: Select the Correct Option

Based on the calculated average evaporation rate, the correct answer corresponds to option (a) \(4.31 \text{ g/min}\).

Key Concepts

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics, as it explains the movement of thermal energy from one object to another. In this context, the child with a fever loses thermal energy to reduce their body temperature from 101°F to 98°F. When the body temperature decreases, it implies there's a transfer of heat from the body to the surroundings. This change occurs via the evaporation of sweat in this scenario.

Heat can be transferred in several ways, such as:

• Conduction: Direct transfer of heat between molecules in contact.
• Convection: Transfer of heat by the movement of fluids (liquids or gases).
• Radiation: Transfer of heat through electromagnetic waves.
• Evaporation: Loss of heat due to liquid turning into vapor.

Understanding heat transfer mechanisms is crucial, particularly in this problem, since evaporation is exclusively assumed to cause the body’s heat loss.

Evaporation

Evaporation is a significant cooling process and involves turning liquid into vapor. In the given exercise, sweat evaporates from the child’s skin to dissipate heat. This process uses the child's body heat to change sweat (liquid) to vapor (gas), subsequently lowering body temperature.

Factors affecting evaporation include:

• Temperature: Higher temperatures increase evaporation rates.
• Surface area: Larger areas facilitate more evaporation.
• Humidity: Lower humidity leads to faster evaporation.
• Wind: Increased air movement can speed up evaporation.

In daily life, evaporation is essential in regulating body temperature and cooling processes. The specific role it plays in the exercise highlights evaporation’s effectiveness in fever reduction.

Specific Heat Capacity

Specific heat capacity refers to the amount of heat required to change the temperature of a unit mass of a substance by 1°C. In the exercise, the specific heat of the human body is approximated as that of water, which is 1 cal/g°C. This property helps us calculate how much heat the child's body loses due to a temperature drop.

The formula to determine heat loss is:\[ Q = m \cdot c \cdot \Delta T \]where

• \(Q\) is the total heat loss.
• \(m\) is the mass.
• \(c\) is the specific heat capacity.
• \(\Delta T\) is the change in temperature.

This relationship shows how specific heat capacity helps us understand the energy exchanges in thermal systems, making it indispensable in thermodynamics.

Latent Heat

Latent heat is the heat absorbed or released during a phase change, such as when a solid turns into a liquid, or a liquid into a gas, without changing temperature. In this exercise, latent heat of evaporation (\(580 \text{ cal/g}\)) is used to calculate how much water evaporates from the child's body.

The formula to find the mass of evaporated water is:\[ m_{\text{water}} = \frac{Q}{L} \]where

• \(Q\) is the heat energy.
• \(L\) is the latent heat.

Latent heat is crucial in problems involving phase transitions, like evaporation, as it quantifies the energy exchanges during the process. Understanding latent heat helps grasp how energy is conserved during phase changes, influencing heat regulation in systems like the human body.