As the steam touches the skin, it undergoes a phase change and releases latent heat due to the phase change. As it reaches equilibrium, it releases sensible heat. We calculate as follows:
Q = latent heat + sensible Heat
Q = 2.26 kJ / g (50.0 g) + 50.0 g ( 4.18 J / g C) (37 C - 100 C) ( 1 kJ / 1000 J)
Q = 99.833 kJ
Answer:
I have no clue sorry I wish I could help
<span>Drought
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Explanation:
Given that,
Radius of the disk, r = 0.25 m
Mass, m = 45.2 kg
Length of the ramp, l = 5.4 m
Angle made by the ramp with horizontal, 
Solution,
As the disk starts from rest from the top of the ramp, the potential energy is equal to the sum of translational kinetic energy and the rotational kinetic energy or by using the law of conservation of energy as :
(a) 
h is the height of the ramp


v is the speed of the disk's center
I is the moment of inertia of the disk,






v = 4.52 m/s
(b) At the bottom of the ramp, the angular speed of the disk is given by :



Hence, this is the required solution.