Answer:
Amount of heat energy released by light bulb = 25 joules
Explanation:
Given:
Energy receive by light bulb = 100 Joules
Energy released by light bulb as light energy = 75 Joules
Find:
Amount of heat energy released by light bulb
Computation:
We know that, energy is neither be created nor destroys
So,
Using Law of conservation of energy
Energy receive by light bulb = Energy released by light bulb as light energy + Amount of heat energy released by light bulb
100 = 75 + Amount of heat energy released by light bulb
Amount of heat energy released by light bulb = 100 - 75
Amount of heat energy released by light bulb = 25 joules
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
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Answer: 15.8 g of 
will be required to produce 1.60 grams of 
Explanation:
To calculate the moles :
According to stoichiometry :
As 1 mole of is given by = 2 moles of
Thus 0.05 moles of is given by =
of
Mass of 
Thus 15.8 g of will be required to produce 1.60 grams of
Answer:
Explanation: The median is the value separating the higher half from the lower half of a data sample. For a data set, it may be thought of as the "middle" value. For example, in the data set {1, 3, 3, 6, 7, 8, 9}, the median is 6, the fourth largest, and also the fourth smallest,
