Problem 41 - Arihant AIEEE Physics

Question

In similar calorimeters, equal volumes of water and alcohol, when poured, take \(100 \mathrm{~s}\). and 74 s respectively to cool from \(50^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\). If the thermal capacity of each calorimeter is numerically equal to volume of either liquid, then the specific heat capacity of alcohol is : (Given: the relative density of alcohol as \(0.8\) and specific heat capacity of water as \(1 \mathrm{cal} / \mathrm{g} /{ }^{\circ} \mathrm{C}\) ) (a) \(0.8 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\) (b) \(0.6 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\) (c) \(0.9 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\) (d) \(1 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

(c) The specific heat capacity of alcohol is approximately \(0.9 \mathrm{cal}/\mathrm{g}^{\circ}C\).

Step 1: Understanding the Problem

We have equal volumes of water and alcohol cooling in similar calorimeters. Time taken for water to cool is 100 seconds, and for alcohol, it is 74 seconds. We need to find the specific heat capacity of alcohol given its relative density and the specific heat capacity of water.

Step 2: Calculate Mass of Liquids

Since the volumes are equal and the relative density of alcohol is 0.8, the mass of water, \(m_w\), is simply \(V\) (in suitable units), and the mass of alcohol, \(m_a\), is \(0.8 \times V\).

Step 3: Energy Lost by Water

For water, the heat lost \(Q_w\) is equal to the mass \(m_w\) times its specific heat capacity \(c_w\) times the temperature change \(\Delta T = 10^{\circ}C\). So, \(Q_w = V \times 1 \times 10 = 10V\) calories.

Step 4: Energy Lost by Alcohol

Similarly, for alcohol, the heat lost \(Q_a\) equals the mass \(m_a\) times its specific heat capacity \(c_a\) times the temperature change. So, \(Q_a = 0.8V \times c_a \times 10 = 8Vc_a\).

Step 5: Applying Dulong-Petit Law

The rate of heat loss is proportional to the temperature difference and the total thermal capacity. Since the calorimeters are similar, and both liquids are cooling from the same initial temperature to the same final temperature, the rate comparison is directly proportional to the specific heats: \(\frac{Q_w}{t_w} = \frac{Q_a}{t_a}\).

Step 6: Substitute Values and Solve

Substitute values and solve the equation: \(\frac{10V}{100} = \frac{8Vc_a}{74}\). Simplifying, \(c_a = \frac{10V \times 74}{8V \times 100} = \frac{740}{800} = 0.925 \approx 0.9 \mathrm{cal}/\mathrm{g}^{\circ}C\).

Step 7: Conclusion

The specific heat capacity of alcohol is approximately \(0.9 \mathrm{cal}/\mathrm{g}^{\circ}C\). The closest match in the given options is (c) \(0.9 \mathrm{cal}/\mathrm{g}^{\circ}C\).

Key Concepts

Calorimetry

Calorimetry is the science of measuring the amount of heat transferred to or from a substance. It plays a crucial role in understanding how substances absorb, transfer, and retain heat.
In this exercise, calorimetry helps us compare the cooling of water and alcohol. Both liquids are placed in calorimeters to observe how they release heat. By measuring the time it takes to cool from one temperature to another, we can derive critical values like specific heat capacity.
Calorimeters ensure an isolated environment, meaning all observed heat exchanges occur within the system. This isolation allows accurate measurements of heat flow, helping us understand thermal properties of substances.

• Calorimeters prevent external heat loss or gain.
• Observing heat exchange helps determine specific heat capacities.
• The experiment illustrates how different substances behave under identical conditions.

Thermal Capacity

Thermal capacity, or heat capacity, refers to the amount of heat needed to change a substance's temperature by a specified amount. It directly relates to the substance's mass and specific heat capacity.
In the given exercise, each calorimeter's thermal capacity is equal to the volume of either liquid. This means the same amount of heat causes different temperature changes in water and alcohol due to their differing thermal capacities.

• Thermal capacity = mass x specific heat capacity.
• It indicates how much heat a substance can absorb or release.
• In the calculation, equal volumes showcase diverse thermal capacities due to differing masses.

This principle explains why different substances change temperature at different rates when subjected to identical conditions.

Relative Density

Relative density, or specific gravity, is the ratio of the density of a substance to the density of a reference material, usually water for liquids.
In the problem, the relative density of alcohol is given as 0.8, meaning alcohol is less dense than water. Knowing this allows us to calculate the mass of alcohol when given its volume.
This concept helps in determining how substances behave under specific conditions, like cooling rates.

• Relative density < 1 indicates a substance is less dense than the reference.
• It affects how we calculate masses from volumes.
• Determines the buoyancy and settling behavior of substances.

Understanding relative density is vital for solving problems where mass and volume relationships are crucial.

Heat Transfer

Heat transfer refers to the movement of thermal energy due to temperature differences. It occurs spontaneously from warmer to cooler objects until thermal equilibrium is reached.
In the exercise, we explore how both water and alcohol release heat as they cool. The rate of heat transfer affects how quickly each liquid reaches the desired temperature.

• Occurs via conduction, convection, and radiation.
• Essential for understanding temperature changes in substances.
• Affects how we experience temperature in the environment.

This process elucidates the physical principles governing temperature changes, and is foundational for solving calorimetry problems.