Question

You have 99 g of steam at 100◦C. The specific heat capacity of water is 4180 J/kg◦C and latent heat of vaporiza- tion of water is 2.26 × 10⁶ J/kg . How much heat must be removed to change it to 99 g of water at 29◦C?

Answer in units of J.

Answer 1

To change 99 g of steam at 100°C to 99 g of water at 29°C, a total of 253,273.04 J of heat must be removed, which includes the latent heat of vaporization and the heat removed to lower the temperature.

Step-by-step explanation:

The student asked how much heat must be removed to change 99 g of steam at 100°C to 99 g of water at 29°C. To solve this, two steps are required: First, the condensation of steam to water, followed by the cooling of the water from 100°C to 29°C. The amount of heat removed during condensation (latent heat) can be calculated with the formula Qv = m x Lv, where m is the mass and Lv is the latent heat of vaporization. The latent heat of vaporization for water is given as 2.26 × 106 J/kg. After converting mass to kg (m = 0.099 kg), we find Qv = 0.099 kg × 2.26 × 106 J/kg = 223,740 J.

For cooling, the formula Q = m x c x ΔT is used, where c is the specific heat capacity and ΔT is the temperature change. Given the specific heat capacity of water is 4.184 J/g°C, we calculate Q = 99 g x 4.184 J/g°C x (100°C - 29°C) = 99 g x 4.184 J/g°C x 71°C = 29,533.04 J. Adding both values together, the total heat to be removed is Qtotal = 223,740 J + 29,533.04 J = 253,273.04 J.