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Condensation is a critical concept in heat transfer calculations. It happens when steam (or any gas) changes into a liquid state. During this process, heat energy, known as latent heat, is released. This heat release can be significant, as seen in the given problem where steam at 100°C condenses to water.
In the example, we need to calculate the heat released when 18.0 grams of steam condenses at 100°C. The formula used is:
\[ q = m \times \text{ΔH}_{vap} \] Here, \( m \) is the mass (18.0 grams) and \( \text{ΔH}_{vap} \) is the heat of vaporization of water (540 cal/g).
Substituting the values, we get:\[ q = 18.0 \text{ g} \times 540 \frac{\text{cal}}{\text{g}} = 9720 \text{ cal} \] Convert calories to kilocalories:\[ 9720 \text{ cal} \times \frac{1 \text{ kcal}}{1000 \text{ cal}} = 9.72 \text{ kcal} \]So when the steam condenses, it releases 9.72 kilocalories.

Specific heat capacity is the amount of heat required to change the temperature of 1 gram of a substance by 1°C. It is crucial for calculating heat transfer when a substance warms up or cools down. In water, this value is particularly significant.
In our problem, once the steam condenses to liquid water, we need to cool this water from 100°C to the body temperature of 37°C. The formula used for this calculation is:
\[ q = m \times c \times \text{ΔT} \]Here, \( m \) is the mass (18.0 grams), \( c \) is the specific heat capacity of water (1 cal/g°C), and \( \text{ΔT} \) is the change in temperature (100°C - 37°C = 63°C).By substituting the values, we get:\[ q = 18.0 \text{ g} \times 1 \frac{\text{cal}}{\text{g} \text{°C}} \times 63°C = 1134 \text{ cal} \]Convert calories to kilocalories:\[ 1134 \text{ cal} \times \frac{1 \text{kcal}}{1000 \text{ cal}} = 1.134 \text{ kcal} \]So, the cooling water releases 1.134 kilocalories.

Temperature change is a frequent factor in heat transfer calculations. When a substance's temperature changes, it either absorbs or releases heat. The total heat transferred depends both on the amount of temperature change and the substance's specific heat capacity.
In the discussed problem, we have a two-step temperature change. Firstly, steam at 100°C condenses to liquid water at 100°C, then this water cools down to 37°C. Here's how we calculated the total heat released:* First part (condensation): \[ 9.72 \text{ kcal} \]* Second part (cooling): \[ 1.134 \text{ kcal} \]Adding both heat values gives us the total heat released:\[ q_{\text{total}} = 9.72 \text{ kcal} + 1.134 \text{ kcal} = 10.854 \text{ kcal} \]Thus, the total heat released when steam condenses and cools to 37°C is 10.854 kilocalories. Understanding this process breaks down complicated problems into manageable steps.