Problem 53 - Arihant AIEEE Physics

Question

If in \(1.1 \mathrm{~kg}\) of water which is contained in a calorimeter of water equivalent \(0.02 \mathrm{~kg}\) at \(15^{\circ} \mathrm{C}\), steam at \(100^{\circ} \mathrm{C}\) is passed, till the temperature of calorimeter and its contents rises to \(80^{\circ} \mathrm{C}\). The mass of steam condensed in kilogram is: (a) \(0.131\) (b) \(0.065\) (c) \(0.260\) (d) \(0.135\)

Step-by-Step Solution

Answer

The mass of steam condensed is approximately 0.135 kg.

Step 1: Identify Known Quantities

First, list all the known quantities in the problem:- Mass of water in calorimeter, \( m_w = 1.1 \) kg - Water equivalent of calorimeter, \( W_c = 0.02 \) kg- Initial temperature \( T_i = 15^{\circ} \mathrm{C} \)- Final temperature \( T_f = 80^{\circ} \mathrm{C} \)- Temperature of steam, \( T_s = 100^{\circ} \mathrm{C} \)- Specific heat capacity of water, \( c_w = 4.18 \) kJ/kg°C- Latent heat of vaporization of steam, \( L = 2260 \) kJ/kg.

Step 2: Calculate Heat Gained by Water

The heat gained by the water in the calorimeter can be calculated using the formula:\[ Q_w = m_w \times c_w \times (T_f - T_i) \]Substitute the known values:\[ Q_w = 1.1 \times 4.18 \times (80 - 15) \]Calculate:\[ Q_w = 1.1 \times 4.18 \times 65 = 299.87 \text{ kJ} \].

Step 3: Calculate Heat Gained by Calorimeter

The heat gained by the calorimeter alone is:\[ Q_c = W_c \times c_w \times (T_f - T_i) \]Substitute the values:\[ Q_c = 0.02 \times 4.18 \times 65 \]Calculate:\[ Q_c = 5.434 \text{ kJ} \].

Step 4: Total Heat Gained

The total heat gained by both water and the calorimeter is the sum of the above two calculations:\[ Q_{\text{total}} = Q_w + Q_c \]\[ Q_{\text{total}} = 299.87 + 5.434 = 305.304 \text{ kJ} \].

Step 5: Calculate Mass of Steam Condensed

The mass of the steam condensed can be determined from the latent heat of vaporization:\[ Q_{\text{steam}} = m_s \times L \]Set \( Q_{\text{total}} = Q_{\text{steam}} \) to find \( m_s \):\[ 305.304 = m_s \times 2260 \]Solve for \( m_s \):\[ m_s = \frac{305.304}{2260} \approx 0.135 \text{ kg} \].

Key Concepts

Calorimetry

Calorimetry focuses on the measurement of heat transfer during physical or chemical processes. It helps us determine how much heat is exchanged between substances at different temperatures, such as when hot steam is condensed in a cooler calorimeter filled with water. In the exercise, the issue is to calculate the amount of steam condensed in water within a calorimeter.
Here, concepts like heat gained by water, or heat gained by the calorimeter, play a role in understanding energy transfer. To comprehend calorimetry, always identify the initial and final temperatures and the specific heat capacity of the involved substances. These factors are vital as they determine the total heat exchange.

Latent Heat of Vaporization

Latent heat of vaporization is the energy required to change a substance from liquid to gas phase without altering its temperature. When steam condenses, it releases this heat. Steam at 100°C changes back to water, transferring latent heat to its surroundings in this way. In our problem, calculating how much steam is condensed uses this principle. The latent heat of vaporization of water is given as 2260 kJ/kg. This value means each kilogram of steam releases 2260 kJ of energy when it condenses into water. By knowing the total heat gained by the system, you can compute the mass of the steam condensed.

Specific Heat Capacity

Specific heat capacity is the amount of heat needed to raise the temperature of one kilogram of a substance by one degree Celsius. For water, this specific heat capacity is 4.18 kJ/kg°C. This value helps us find out how much heat is absorbed by water when its temperature changes.
In the exercise, the water's temperature increased from 15°C to 80°C. By using the specific heat capacity, you can calculate how much heat was added to achieve this rise. This concept is crucial in calorimetry problems as it tells us how efficiently a material can absorb heat.

Temperature Change

Temperature change measures how much a substance's temperature increases or decreases when heat is transferred.
This concept is straightforward yet crucial in calorimetry because it helps calculate heat exchange.For example, in the problem, the water and calorimeter initially sat at 15°C and reached 80°C, a notable rise in temperature. By knowing the initial and final temperatures, you can use the formula _\( Q = mc\Delta T \)_ to evaluate the heat gained or lost, where \( \Delta T \) (the temperature change) is a necessary factor.
Always ensure to use consistent units while dealing with temperature changes.

Mass of Water

The mass of water is an essential factor in heat transfer calculations because it directly influences the amount of heat absorbed or released. In many calorimetry problems, the mass of water determines how much heat it can absorb as its temperature changes. In the exercise, you have 1.1 kg of water in the calorimeter. You also need to consider the water equivalent of the calorimeter itself, representing how much additional water could absorb heat comparably. This was given as 0.02 kg in the problem. Combining these values helps to correctly account for total heat gain or loss in the system, leading to a better understanding of energy transfer during the process.