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
<em>Chemists use the mole unit to represent 6.022 × 10 23 things, whether the things are atoms of elements or molecules of compounds. This number, called Avogadro's number, is important because this number of atoms or molecules has the same mass in grams as one atom or molecule has in atomic mass units. </em>
hope helpful~
You are correct, but you needn't worry about the signs so much. Just remember that the negative sign is used to denote a loss of energy; since the water is hotter, it will be losing energy (-Q) and the iron will gain energy (Q). Now, we substitute the values:
-149.3 * 4.184 * (T - 95) = 412 * 0.44 * (T - 5)
Solving this equation for T,
T = 74.8 °C
Answer:
elements are composed of
electrons with a little mass and negative charge
protons with mass and positive charge
neutrons with mass and no charge
isotopes same number of protons, different in neutrons
share me the other part of the photo to complete!!!
