Answer:
The heat transferred from water to skin is 6913.5 J.
Explanation:
Given that,
Weight of water = 25.0 g
Suppose 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=4190 J/(kg°K) for both liquid water and steam.
We need to calculate the heat transferred from water to skin
Using formula for stream
Put the value into the formula
Hence, The heat transferred from water to skin is 6913.5 J.
Explanation:
The given data is as follows.
Mass of small bucket (m) = 4 kg
Mass of big bucket (M) = 12 kg
Initial velocity () = 0 m/s
Final velocity () = ?
Height = 2 m
and, = 0 m
Now, according to the law of conservation of energy
starting conditions = final conditions
235.44 = + 78.48
= 4.43 m/s
Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.
As per kinematics equation we are given that
now we are given that
a = 2.55 m/s^2
now we need to find x
from above equation we have
so it will cover a distance of 93.2 m