In a steam radiator, steam at 110°C condenses and the liquid water formed is cooled to 80°C. How much heat per gram of liquid formed, is released in this process?
To calculate the heat released per gram of liquid formed in this process, we need to use the specific heat capacity of water and the heat of vaporization.
First, let's calculate the heat released during the cooling of steam to liquid water:
Q1 = m * c * ΔT1
Where:
Q1 is the heat released during cooling (in joules)
m is the mass of steam that condenses (in grams)
c is the specific heat capacity of water (4.186 J/g°C)
ΔT1 is the change in temperature during cooling (from 110°C to 80°C)
Next, let's calculate the heat released during the condensation of steam to liquid water:
Q2 = m * ΔHvap
Where:
Q2 is the heat released during condensation (in joules)
m is the mass of steam that condenses (in grams)
ΔHvap is the heat of vaporization of water (2260 J/g)
Finally, let's find the total heat released during the process:
Total heat released = Q1 + Q2
Now, let's plug in the values and calculate:
ΔT1 = 80°C - 110°C = -30°C
ΔHvap = 2260 J/g
Q1 = m * c * ΔT1
Q2 = m * ΔHvap
Total heat released = Q1 + Q2
To find the heat released per gram of liquid formed, we divide the total heat released by the mass of water formed:
Heat released per gram = (Q1 + Q2) / m
Please provide the mass of steam that condenses, and I'll be able to calculate the answer for you.