How much heat is absorbed by a persons hand if 100 grams of steam at 100°C is first converted to liquid water at 100°C and then cooled to normal body temperature?
q1 = heat removed to condense steam at 100 to liquid at 100 C.
q1 = mass x heat vaporization
q2 = heat removed to cool from liquid water at 100 C to body temperature of 37 C.
q2 = mass water x specific heat water x (Tfinal-Tinitial). Tf = 37, Ti = 100.
Total heat absorbed by hand is q1 + q2.
To calculate the heat absorbed by a person's hand in this scenario, we need to take into account two processes: first, the heat required to condense the steam into liquid water, and second, the heat required to lower the temperature of that liquid water to body temperature (assumed to be around 37°C).
1. Heat absorbed during condensation:
The heat absorbed during condensation can be calculated using the formula Q = m * ΔH, where Q is the heat absorbed, m is the mass of the substance, and ΔH is the heat of condensation.
Given:
Mass of steam (m) = 100 grams
Heat of condensation (ΔH) = 2260 J/g (at atmospheric pressure)
Q = 100 g * 2260 J/g = 226,000 J
Therefore, 226,000 joules (J) of heat are absorbed during the condensation process.
2. Heat absorbed during cooling:
To calculate the heat absorbed during the cooling process, we need to use the specific heat capacity formula Q = m * c * ΔT, where Q is the heat absorbed, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
Since we want to cool the liquid water from 100°C to 37°C, ΔT = 100°C - 37°C = 63°C.
Given:
Mass of water (m) = 100 grams (conserved during condensation)
Specific heat capacity of water (c) = 4.18 J/g°C
Q = 100 g * 4.18 J/g°C * 63°C = 26,334 J
Therefore, 26,334 joules (J) of heat are absorbed during the cooling process.
Total heat absorbed = Q(condensation) + Q(cooling)
Total heat absorbed = 226,000 J + 26,334 J
Total heat absorbed ≈ 252,334 J
Therefore, approximately 252,334 joules (J) of heat are absorbed by a person's hand in this scenario.