Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :
where,
= change in entropy
= change in enthalpy of vaporization = 40.5 kJ/mol
= boiling point temperature = 352 K
Now put all the given values in the above formula, we get:
Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C
<u>Explanation:</u>
When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.
The equation used to calculate heat released or absorbed follows:
......(1)
where,
q = heat absorbed or released
= mass of hot water = 46.7 g
= mass of cold water = 45.33 g
= final temperature = 59.4°C
= initial temperature of hot water = 80.6°C
= initial temperature of cold water = 40.6°C
= specific heat of hot water = 4.184 J/g°C
= specific heat of cold water = 4.184 J/g°C
= specific heat of calorimeter = ? J/g°C
Putting values in equation 1, we get:
Hence, the specific heat of calorimeter is 30.68 J/g°C