Answer : The final temperature of the mixture is 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
And as we know that,
Mass = Density × Volume
Thus, the formula becomes,
where,
= specific heat of ethanol = 
= specific heat of water = 
= mass of ethanol
= mass of water
= density of ethanol = 0.789 g/mL
= density of water = 1.0 g/mL
= volume of ethanol = 45.0 mL
= volume of water = 45.0 mL
= final temperature of mixture = ?
= initial temperature of ethanol = 
= initial temperature of water = 
Now put all the given values in the above formula, we get
Therefore, the final temperature of the mixture is
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Answer: 100C of heat is needed.
Explanation: That is the heating point of water
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
Explanation:
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