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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. We will further assume a constant specific heat capacity 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

Respuesta :

Answer:

63.3 KJ

Explanation:

Quantity of heat lost to the skin = mLw + mcΔT

where is mass in kg, Lw is latent heat of vaporization in KJ/kg ( 2.256 × 10^6 KJ/Kg), c is specific heat capacity of water (4190 J/(Kg.K) and ΔT is the change in temperature

m = 25g = 25 /1000 = 0.025 kg

quantity of heat loss = (0.025 × 2.256 × 10^6) +  ( 0.025 × 4190 × (100 -34 )

Quantity of heat loss to the skin = (5.64 × 10^4) + 5530.8 = 56400 + 6913.5 = 63313.5 = 63.3 KJ

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