Find the amount of heat to be released to convert 40 grams of steam at 108°C to ice at -7°C. The latent heat of fusion and the latent heat of vaporization are 80 cal/g and 540 cal/g respectively. Cp water 4.187 KJ/kg K, Cp ice 2.108 KJ/kg.K, Cp steam 1.996 KJ/kg.K
To find the amount of heat released in this process, we need to break it down into three steps:
1. Heating the steam from 108°C to 100°C
2. Condensing the steam into water at 100°C
3. Cooling the water from 100°C to -7°C and solidifying it into ice
Let's calculate the heat required for each step:
Step 1: Heating the steam from 108°C to 100°C
Since we're dealing with grams instead of kilograms, first convert the mass of steam to kilograms:
Mass of steam = 40 g = 0.04 kg
To find the heat required to raise the temperature of the steam, we'll use the equation:
Q = m * Cp * ΔT
where Q is the heat required in Joules, m is the mass in kilograms, Cp is the specific heat capacity in kJ/kg·K, and ΔT is the change in temperature in Kelvin.
The change in temperature is 100°C - 108°C = -8°C, which is -8 K because the temperature is in Celsius.
Q1 = 0.04 kg * 1.996 kJ/kg·K * (-8 K) = -0.638 kJ
Step 2: Condensing the steam into water at 100°C
The heat required to condense the steam can be calculated using the equation:
Q2 = m * L_v
where L_v is the latent heat of vaporization.
Q2 = 0.04 kg * 540 cal/g * (4.187 kJ/cal) = 8.8064 kJ
Step 3: Cooling the water from 100°C to -7°C and solidifying it into ice
First, calculate the heat required to cool the water from 100°C to 0°C:
Q3a = m * Cp * ΔT
Q3a = 0.04 kg * 4.187 kJ/kg·K * (-100 K) = -16.748 kJ
Then, calculate the heat required to freeze the water into ice at 0°C:
Q3b = m * L_f
where L_f is the latent heat of fusion.
Q3b = 0.04 kg * 80 cal/g * (4.187 kJ/cal) = 1.33856 kJ
Finally, calculate the heat required to cool the ice from 0°C to -7°C:
Q3c = m * Cp * ΔT
Q3c = 0.04 kg * 2.108 kJ/kg·K * (-7 K) = -0.58768 kJ
Now, add up the heat from each step to find the total heat released:
Total heat released = Q1 + Q2 + Q3a + Q3b + Q3c
Total heat released = -0.638 kJ + 8.8064 kJ - 16.748 kJ + 1.33856 kJ - 0.58768 kJ = -7.82872 kJ
Therefore, the amount of heat released to convert 40 grams of steam at 108°C to ice at -7°C is approximately 7.83 kJ (rounded to two decimal places).