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In any calorimetry problem, understanding heat exchange is essential. Heat exchange describes the process where thermal energy is transferred from one substance to another. This transfer occurs until the entire system reaches thermal equilibrium. In the given exercise, steam loses heat as it condenses and then cools down, while ice and water gain heat to increase their temperature.
Importantly, no heat is lost to the surroundings in a perfectly insulated system like an ideal calorimeter. This means the heat given off by the steam exactly equals the heat absorbed by the ice and water. This balance of energy is crucial in solving calorimetry problems as it helps us to set up equations to find the unknowns, such as the final temperature, in this scenario.

Thermal equilibrium is reached when all parts of a system have the same temperature. At this point, there is no net flow of thermal energy—heat. For calorimetry exercises, understanding this concept helps us know when to stop calculations: it’s when the temperature stops changing. In our scenario, after steam condenses and releases heat, and ice absorbs heat to melt, the entire mixture will eventually balance out at a uniform final temperature.
Reaching thermal equilibrium tells us that all energy exchanges have completed. This principle guides the creation of the primary equation used to determine final states of temperature in calorimetry problems. The heat lost by the hot component must equal the heat gained by the cooler components.

Latent heat is the energy absorbed or released by a substance during a phase change without changing its temperature. There are two types in calorimetry: latent heat of fusion and latent heat of vaporization.
The latent heat of fusion is the energy needed to change ice to water at 0°C without changing temperature. For this exercise, the amount of heat absorbed by the ice during melting is calculated using the formula: \[ Q = m \times L_f \]where \( m \) is the mass and \( L_f \) is the latent heat of fusion.
The latent heat of vaporization is the heat required to transform steam into liquid water, again with no temperature change. In our scenario, when the steam condenses at 100°C, it releases a significant amount of heat, utilizing the latent heat: \[ Q = m \times L_v \]where \( m \) is mass and \( L_v \) is latent heat of vaporization. Recognizing these concepts enables you to predict how much heat is involved during phase changes in calorimetry.

Specific heat capacity is a material's ability to absorb energy in the form of heat. Specifically, it's the amount of heat required to raise the temperature of 1 gram of the substance by 1°C. For water, a common substance studied in calorimetry, the specific heat capacity is 4.18 J/g°C.
In calorimetry problems, like our step-by-step solution, specific heat capacity helps calculate how much heat is absorbed or released as substances change temperature. We use this value to determine the heat needed to raise the melted ice, initial water, and calorimeter equivalent to the final temperature.
The formula utilized is:\[ Q = m \times c \times \Delta T \]where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature. Understanding this principle allows us to calculate and predict the behavior of different substances in heat exchanges efficiently.