Question

\(100 \mathrm{~g}\) ice at \(0^{\circ} \mathrm{C}\) placed in \(100 \mathrm{~g}\) water at \(100^{\circ} \mathrm{C}\). The final temperature of the mixture will be........ (Latent heat of ice is \(80 \mathrm{Cal} / \mathrm{g}\), and specific heat of water is \(1 \mathrm{Cal} / \mathrm{g} \mathrm{C}^{\circ}\) ) (A) \(10^{\circ} \mathrm{C}\) (B) \(20^{\circ} \mathrm{C}\) (C) \(30^{\circ} \mathrm{C}\) (D) \(50^{\circ} \mathrm{C}\)

Short answer

The final temperature of the mixture is (B) \(20^{\circ} \mathrm{C}\).

Step-by-step solution

Write down the heat gained by ice and heat lost by water

To calculate the final temperature, we need to find the heat gained by ice when it turns into water and the heat lost by water when it cools down in the process. The heat transfer process can be described with the following equation: Heat gained by ice = Heat lost by water

Express heat gained by ice in terms of mass, latent heat, and specific heat of water

When the ice turns into water at \(0^{\circ} \mathrm{C}\), it will gain heat due to its latent heat and then gain more heat to reach the final temperature. So, the heat equation of ice becomes: \( Q_{ice} = m_{ice} \times L + m_{ice} \times c \times (T_f - 0) \) Here, \(m_{ice}\) - mass of ice (\(100 \mathrm{g}\)) \(\ L\) - latent heat of ice (\(80 \mathrm{Cal} / \mathrm{g}\)) \(c\) - specific heat of water (\(1 \mathrm{Cal} / \mathrm{g} \mathrm{C}^{\circ}\)) \(T_f\) - final temperature

Express heat lost by water in terms of mass, specific heat, and final temperature

The water's heat equation when cooling down from \(100^{\circ} \mathrm{C}\) to the final temperature is: \( Q_{water} = m_{water} \times c \times (100 - T_f) \) Here, \(m_{water}\) - mass of water (\(100 \mathrm{g}\))

Set the heat gained by ice equal to the heat lost by water and solve for the final temperature

Using the principle of conservation of energy, set the heat gained by ice equation equal to the heat lost by water equation and solve for the final temperature: \( m_{ice} \times L + m_{ice} \times c \times (T_f - 0) = m_{water} \times c \times (100 - T_f) \) Now, plug in the given values: \( 100 \times 80 + 100 \times 1 \times (T_f - 0) = 100 \times 1 \times (100 - T_f) \) By solving the above equation, we get the final temperature: \( T_f = 20^{\circ} \mathrm{C} \) So, the answer is (B) \(20^{\circ} \mathrm{C}\).

Key concepts

Latent Heat

Latent heat is an essential concept in thermodynamics explaining the energy required for a substance to change its phase without affecting its temperature. When you melt ice to water or boil water to steam, energy goes into breaking the bonds that hold molecules in a solid or liquid form. Although this energy is absorbed or released, the temperature remains constant during the phase change.

The latent heat of fusion, for instance, refers to the heat required to turn ice into water at 0°C. In our exercise, each gram of ice absorbs 80 calories (latent heat of ice) to convert from solid to liquid, without raising the temperature. This process allows us to calculate the amount of heat needed for complete phase transition, crucial for understanding thermal interactions like those between ice and hot water.

Specific Heat Capacity

The specific heat capacity is a measure of how much heat energy is required to raise the temperature of a given mass of a substance by 1 degree Celsius. Different substances have unique specific heat capacities, reflecting their ability to store or absorb heat energy. For example, water has a high specific heat capacity of 1 Cal/g°C, meaning it requires more energy to change temperature compared to other substances.

Understanding specific heat capacity is vital in the exercise to determine how much heat the ice and water can absorb or release. When water cools from 100°C, it releases heat at a rate that is determined by its specific heat. This aids in calculating energy exchange with the ice, considering the temperatures and masses involved.

Heat Transfer

Heat transfer is the flow of thermal energy from a hotter object to a cooler one. In the context of the exercise, it describes the energy exchange that happens between the ice and the hot water until they reach thermal equilibrium. There are three main modes of heat transfer: conduction, convection, and radiation, but in our exercise, the primary mode is conduction.

During the mixing of ice and hot water, heat transfers from the warm water to the cooler ice. This transfer causes the ice to melt and the water to cool down until both reach the same temperature. Understanding how heat moves between substances allows us to predict the final temperature accurately, as seen when solving for thermal equilibrium in this task.

Calorimetry

Calorimetry is the science of measuring heat based on observing temperature changes. It is a crucial tool in thermodynamics for analyzing heat interactions within chemical reactions or physical changes like phase transitions. In the given exercise, calorimetry principles are applied by ensuring energy conservation through the heat exchange between ice and water.

Calorimetry involves calculating and equating the heat gained by the ice to the heat lost by the water, ensuring the law of conservation of energy holds. By considering both latent heat and specific heat capacities, we can use calorimetry to determine the final equilibrium temperature. This incorporates all elements of the calculations used, providing a complete analysis of the thermal interaction process in the mix of different temperatures.