Question

Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat input to your skin from steam as opposed to hot water at the same temperature. Assume that water and steam, initially at 100∘C, are cooled down to skin temperature, 34∘C when they come in contact with your skin. Assume that the steam condenses extremely fast. The heat capacity of liquid water is c=4190J/(kg⋅K) for both liquid water and steam.

How much heat H1 is transferred to the skin by 25.0 g of steam onto the skin? The latent heat of vaporization for steam is L=2.256

Answer 1

The heat transferred to the skin by 25.0 g of steam is 279,156.4 J.

Step-by-step explanation:

To calculate the heat transferred to the skin by steam, we need to consider the heat absorbed during phase change and heating up from 34°C to 100°C. First, let's calculate the heat absorbed during phase change. The latent heat of vaporization for steam is 2.256 J/g, so for 25.0 g of steam, the heat absorbed is 25.0 g x 2.256 J/g = 56.4 J.

Next, let's calculate the heat absorbed during heating up from 34°C to 100°C. The heat capacity of liquid water is 4190 J/(g⋅K), so the heat absorbed is (100°C - 34°C) x 25.0 g x 4190 J/(g⋅K) = 279,100 J.

Finally, we can add the two values to get the total heat transferred to the skin by steam, which is 56.4 J + 279,100 J = 279,156.4 J.

Answer 2

H1 = 63.3 kJ

Step-by-step explanation:

Given:

ΔL = 2.256 kJ/g

Cp = 4190J/kg⋅K

Steam at 100°C in contact with your skin, condenses to water at 100°C. Therefore,

Q = M × ΔL

= (25.0 g)(2.256 kJ/g)

= 56.4 kJ

This condensed water at 100°C is then cooled to 34°C. The heat involved with this temperature change is;

Q = m × Cp × ΔT

= (25 × 1kg/1000g × 4190 × (100°C - 34°C)

= 6913.5 J

= 6.914 kJ

The total heat H1,

= (56.4 kJ) + (6.9135 kJ)

= 63.3 kJ.