Answer:
0.027 litres
Explanation:
volume of cube = length × base area
volume of cube = 0.03m ×( 0.03m × 0.03m )
volume of cube = 0.03m × ( 0.0009m^2 )
volume of cube = 0.000027m^3
1 cubic metre = 1000 litres
0.000027m^3 = 0.027 litres
Answer:
The particles in a solid are tightly packed and locked in place. ... The particles in a liquid are close together (touching) but they are able to move/slide/flow past each other. The particles in a gas are fast moving and are able to spread apart from each other.
Explanation:
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
Find it on google i’m pretty sure i saw it somewhere so sorry this doesn’t help
Answer:
ΔT = 0.78 °C
Explanation:
Given data:
Mass of Al = 9.5 g
Specific heat capacity of Al = 0.9 J/g.°C
Temperature change = ?
Heat added = 67 J
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
67 J = 9.5 g × 0.9 j/g.°C × ΔT
67 J = 85.5 j/°C × ΔT
ΔT = 67 J / 85.5 j/°C
ΔT = 0.78 °C
