How much heat is absorbed by a persons hand if 100 grams of steam at 100c is first converted to liquid at 100c and then cooled to normal body temperature?
To determine the amount of heat absorbed by a person's hand, we need to consider the following steps:
Step 1: Calculating the heat required to convert steam to liquid (latent heat of vaporization):
The heat required to convert 100 grams of steam at 100°C to liquid at 100°C can be calculated using the formula:
Q = m * L
Where:
Q is the heat energy
m is the mass of the substance
L is the latent heat of vaporization
The latent heat of vaporization for water is approximately 2260 J/g.
Q = 100 g * 2260 J/g
Q = 226,000 J
So, 226,000 J of heat is required to convert 100 grams of steam at 100°C to liquid at 100°C.
Step 2: Calculating the heat required to cool the liquid to body temperature:
To cool the liquid water to normal body temperature, we need to calculate the heat energy using the formula:
Q = m * c * ΔT
Where:
Q is the heat energy
m is the mass of the substance
c is the specific heat capacity of water
ΔT is the change in temperature
The specific heat capacity of water is approximately 4.18 J/g°C.
Assuming normal body temperature is 37°C, the change in temperature (ΔT) would be 100°C - 37°C = 63°C.
Q = 100 g * 4.18 J/g°C * 63°C
Q = 264,594 J
So, 264,594 J of heat is required to cool the liquid water to normal body temperature.
Therefore, the total heat absorbed by a person's hand would be the sum of the heat required to convert steam to liquid and the heat required to cool the liquid to body temperature.
Total heat absorbed = 226,000 J + 264,594 J
Total heat absorbed = 490,594 J
Therefore, the person's hand would absorb approximately 490,594 Joules of heat.
To calculate the amount of heat absorbed by a person's hand, we need to consider two separate processes: the phase change from steam to liquid and the cooling of the liquid to body temperature.
1. First, let's calculate the heat absorbed during the phase change from steam to liquid.
To convert 100 grams of steam at 100°C to liquid at 100°C, we need to calculate the heat absorbed during the phase change (latent heat of vaporization). The latent heat of vaporization for water is approximately 2260 joules/gram.
Heat absorbed during phase change = Mass of steam * Latent heat of vaporization
= 100 grams * 2260 J/gram
= 226,000 Joules
2. Now, let's calculate the heat absorbed during the cooling process from liquid at 100°C to normal body temperature (assuming body temperature is around 37°C).
The specific heat capacity of water is approximately 4.18 joules/gram°C.
The change in temperature is 100°C - 37°C = 63°C.
Heat absorbed during cooling = Mass of liquid * Specific heat capacity * Change in temperature
Since we already converted 100 grams of steam to a liquid form, the mass remains the same.
Heat absorbed during cooling = 100 grams * 4.18 J/gram°C * 63°C
≈ 26,382 Joules
Therefore, the total heat absorbed by a person's hand is the sum of the heat absorbed during the phase change and the heat absorbed during cooling:
Total heat absorbed = Heat absorbed during phase change + Heat absorbed during cooling
= 226,000 Joules + 26,382 Joules
≈ 252,382 Joules
Please note that this is a simplified calculation and does not take into account factors such as heat loss to the surroundings or the specific heat capacity of the human hand.