A quantity of steam (600 g) at 112°C is condensed, and the resulting water is frozen into ice at 0°C. How much heat was removed?
___kcal
To calculate the heat removed, we need to consider the changes in temperature for both the steam and water.
First, let's calculate the heat required to cool down the steam from 112°C to 100°C. To do this, we can use the formula:
Q = m * c * ΔT
Where:
Q is the heat absorbed/released
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
For steam, the specific heat capacity is approximately 2.03 kcal/g°C.
ΔT = 100°C - 112°C = -12°C (negative because the temperature is decreasing)
Q1 = 600 g * 2.03 kcal/g°C * -12°C
Next, let's calculate the heat required to cool down the water from 100°C to 0°C. The specific heat capacity of water is approximately 1 kcal/g°C.
ΔT = 0°C - 100°C = -100°C
Q2 = 600 g * 1 kcal/g°C * -100°C
Finally, let's calculate the heat required to freeze the water at 0°C into ice at 0°C. The heat of fusion (latent heat) for water is 333.55 kcal/g.
Q3 = 600 g * 333.55 kcal/g
Now, we can calculate the total heat removed by summing up the three quantities:
Total heat removed = Q1 + Q2 + Q3
Simply calculate the sum to get the final answer in kcal.