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Mrrafil [7]
3 years ago
12

Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat input to your s

kin 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
Physics
1 answer:
gtnhenbr [62]3 years ago
3 0

Answer:

H1 = 63.3 kJ

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.

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Answer:

de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m

Explanation:

Given;

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de Broglie wavelength of the bullet is given by;

λ = h / mv

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h is Planck's constant = 6.626 x 10⁻³⁴ J/s

λ = h / mv

λ = (6.626 x 10⁻³⁴ ) / (0.028 x 765)

λ = 3.093 x 10⁻³⁵ m

Therefore, de Broglie wavelength of the bullet is 3.093 x 10⁻³⁵ m

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