How many kilocalories to condense 82g of steam at 100 degress C, cool the liquid and freeze it at 1 degress C.

To calculate the number of kilocalories required to condense steam at 100 degrees Celsius, cool the resulting liquid, and freeze it at 1 degree Celsius, we need to consider several steps. Let's break it down:

1. Heat energy to condense steam: The first step is to calculate the heat energy required to condense the steam. To do this, we can use the formula:

Q = m * L
where Q is the heat energy, m is the mass of the substance, and L is the latent heat of vaporization for water.

The latent heat of vaporization for water is approximately 540 kcal/kg (kcal per kilogram).

Since we have 82 grams of steam, we need to convert it to kilograms by dividing it by 1000:

m = 82 g / 1000 g/kg = 0.082 kg

Now we can calculate the heat energy required to condense the steam:

Q1 = m * L = 0.082 kg * 540 kcal/kg = 44.28 kcal

Therefore, it requires approximately 44.28 kilocalories to condense 82 grams of steam at 100 degrees Celsius.

2. Heat energy to cool the liquid: The next step is to calculate the heat energy required to cool the liquid from 100 degrees Celsius to 1 degree Celsius. To do this, we can use the specific heat capacity formula:

Q = m * c * ΔT
where Q is the heat energy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the temperature change.

The specific heat capacity of water is approximately 1 kcal/kg°C.

Since we have already condensed the steam, we now have liquid water to cool. The mass of the liquid water is still 82 grams, or 0.082 kg.

The temperature change from 100 degrees Celsius to 1 degree Celsius is:

ΔT2 = 100°C - 1°C = 99°C

Now we can calculate the heat energy required to cool the liquid:

Q2 = m * c * ΔT = 0.082 kg * 1 kcal/kg°C * 99°C = 8.118 kcal

Therefore, it requires approximately 8.118 kilocalories to cool the liquid from 100 degrees Celsius to 1 degree Celsius.

3. Heat energy to freeze the liquid: The final step is to calculate the heat energy required to freeze the liquid at 1 degree Celsius. To do this, we consider the latent heat of fusion for water.

The latent heat of fusion for water is approximately 80 kcal/kg.

We have already established that the mass of the liquid water is 0.082 kg.

Now we can calculate the heat energy required to freeze the liquid:

Q3 = m * L_fusion = 0.082 kg * 80 kcal/kg = 6.56 kcal

Therefore, it requires approximately 6.56 kilocalories to freeze 82 grams of water at 1 degree Celsius.

To find the total number of kilocalories required, we sum up the heat energies from each step:

Total kilocalories = Q1 + Q2 + Q3
= 44.28 kcal + 8.118 kcal + 6.56 kcal
= 58.958 kcal

Therefore, it requires approximately 58.958 kilocalories to condense 82 grams of steam at 100 degrees Celsius, cool the liquid, and freeze it at 1 degree Celsius.