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:
3.99998575239 pounds per gallon (US)
Explanation:
A) particles are close together in random positions with about equal kinetic energy and intermolecular forces.
These points are about liquid state.
B) particles are close together in fixed positions with low kinetic energy
These points satisfy the qualities of Solid state
C)particles are far apart with greater kinetic energy and low intermolecular forces.
The above qualities are for Gaseous state of matter
A) Liquid
B)Solid
C)Gas
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3.5 M has 3.5 moles per litre
so we have one litre, so we need 3.5 moles
moles = mass/molarmass
3.5 * 23 = 80.5
