Mk i will comment so u can make them brainliest lols
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
Well, if u had a spilled liquid in there (we'll simply go with water) and you had the freezer at a cold temperature it would change (like,icycles on trees when it's snowing)
Answer:
Pressure = 4313.43mmHg
Explanation:
P1 = ?
V1 = 0.335L
V2 = 1700mL =1700*10^-3L = 1.7L
P2 = 850mmhg
From Boyle's law, the volume of a fixed mass of gas is inversely proportional to its pressure provided that temperature remains constant.
P = k / v
K = pv. P1V1 = P2V2 = P3V3 =........=PnVn
P1V1 = P2V2
Solve for P1,
P1 = (P2*V2) / V1
P1 = (850 * 1.7) / 0.335
P1 = 4313.43mmHg
The pressure of the gas was 4313.43mmHg
