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:
The answer to the question is
The specific heat capacity of the alloy = 1.77 J/(g·°C)
Explanation:
To solve this, we list out the given variables thus
Mass of alloy = 45 g
Initial temperature of the alloy = 25 °C
Final temperature of the alloy = 37 °C
Heat absorbed by the alloy = 956 J
Thus we have
ΔH = m·c·(T₂ - T₁) where ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C and m = mass of the alloy = 45 g
∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c or
c = 956 J/(540 g·°C) = 1.77 J/(g·°C)
The specific heat capacity of the alloy is 1.77 J/(g·°C)
I’m don’t kill me if I’m wrong but I think it’s high melting point
Mass is the property of a physical body and the resistance to acceleration when a net force is applied on the body.
The atomic mass of sodium (Na) is = 22.98
The atomic mass of nitrate (N) is = 14.00
The atomic mass of oxygen (O) is = 15.99
The sodium nitrate (NaNO3) consists of the atomic masses of Na+N+(O)3 = 85 grams
Therefore, the mass of 6.5 mol of sodium nitrate is = 6.5 * 1 mol of NaNO3
= 6.5 * (85)
= 552.50 grams
