Final answer:
The total heat released when steam condenses and then cools from 110°C to 80°C is calculated in two parts: condensation and cooling. For condensation, the latent heat of vaporization at 100°C (539 cal/g) is used. The subsequent cooling is calculated using water's specific heat capacity (1 cal/g°C), yielding a total of 569 cal/g per gram of liquid formed.
Step-by-step explanation:
To calculate the heat released when steam condenses to liquid water and then cools, we need to consider two processes: the condensation of steam and the subsequent cooling of the water from 110°C to 80°C. The amount of heat released during the condensation of steam into water at the same temperature (100°C) is known as the latent heat of vaporization, which is 539 cal/g. Although the question states the initial temperature of steam is 110°C, we will use the latent heat value for 100°C as a close approximation since latent heat of vaporization is not strongly dependent on temperature within a small range. The specific heat capacity of water is approximately 1 cal/g°C which is the amount of heat required to change the temperature of one gram of water by one degree Celsius.
To find the total heat released, we will add the heat released from condensation at 110°C and the heat released while the water cools from 110°C to 80°C. Since we don't have a specific value for the latent heat at 110°C, we'll proceed with the value at 100°C as:
- Heat released during condensation: Q1 = mass × latent heat of vaporization
- = m × 539 cal/g
- Heat released during cooling: Q2 = mass × specific heat capacity × (ΔT)
- = m × 1 cal/g°C × (110°C - 80°C)
The total heat released per gram = Q1 + Q2 thus total heat released = (m × 539 cal/g) + (m × 1 cal/g°C × 30°C).
When we express this on a per-gram basis, the mass (m) is simply 1 gram:
Total heat released per gram of liquid formed = 539 cal/g + 30 cal/g
= 569 cal/g.