The answer is false methane is emitted from coal
Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)
Answer:
-3670.33 J/K
Explanation:
Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.
Mathematically, change of Entropy can be expressed as,
ΔS = ΔH/T ....................................... Equation 1
Where ΔS = Change of entropy, ΔH = heat change, T = temperature.
ΔH = -(Lf×m).................................... Equation 2
Note: ΔH is negative because heat is lost.
Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg
Substitute into equation
ΔH = -(3.34×10⁵×3.0)
ΔH = - 1002000 J.
But T = 0 °C = (0+273) K = 273 K.
Substitute into equation 1
ΔS = -1002000/273
ΔS = -3670.33 J/K
Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C
Answer ( 3 ) :
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hope this helps!
Answer:
Heat required = 13,325 calories or 55.75 KJ.
Explanation:
To convert a water to steam at 100 degree celsius to vapor, we have to give latent heat of vaporization to water
Which equals ,
Q = mL,
Where, m is the mass of water present
L = specific latent heat of vaporization
Here , m= 25 gram
L equals to 533 calories (or 2230 Joules)
So, Q = 25×533 = 13,325 Calories
Or , Q = 55,750 Joules = 55.75 KJ
so, Heat required = 13,325 calories or 55.75 KJ.
Answer:
Explanation:
2 moles hydrogen reacts with one mole of oxygen to give 2 moles of water.
a ) rate of consumption of hydrogen ( moles per second) is twice the rate of consumption of oxygen .
b ) rate of formation of water ( moles per second ) is twice the rate of consumption of oxygen
c ) rate of formation of water ( moles per second ) is equal to the rate of consumption of hydrogen.
