What is the entropy change for freezing 2.71 g of c2h5oh at 158.7 k? ∆h = −4600 j/mol. answer in units of j/k.

-1.71 J/K

Explanation:

To solve this problem we use the formula

ΔS = n*ΔH/T

Where n is mol, ΔH is enthalpy and T is temperature.

ΔH and T are already given by the problem, so now we calculate n:

Molar Mass C₂H₅OH = 46 g/mol

2.71 g C₂H₅OH ÷ 46g/mol = 0.0589 mol

Now we calculate ΔS:

ΔS = 0.0589 mol * −4600 J/mol / 158.7 K

ΔS = -1.71 J/K

the answer is c i believe

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.