The molar heat of vaporization of liquid nitrogen is 5460 J/mol or 5.46 x 10^3 kJ/mol.
Further Explanation:
This problem is an example of a calorimetry problem. In such calorimetry problems, the amount of heat lost (gained) by the reaction is equal to the amount of heat gained (lost) by water in the calorimeter. In this case,
[tex]-q_{N_{2}} \ = q_{H_{2}O[/tex]
To get the value of q, the equation below is used:
[tex]q \ = mC(T_{f} \ - T_{i})[/tex]
where:
q is the hear absorbed
C is the specific heat capacity
T(f) is the final temperature
T(i) is the initial temperature
In the problem, the mass, initial, and final temperature of water are given. It is also known that the specific heat capacity of water is 4.18 J/g-°C. These can be used to solve for q of water:
m = 2.00 x 10^2 g
T(f) = 41.0 °C
T(i) = 48.0 °C
C = 4.18 J/g-°C.
[tex]q_{H_{2}O} \ = (2.00 \ x \ 10^2 \ g)(4.18 \ \frac{J}{g- \°C})(41.0 \°C \ - 48.0 \°C)\\\boxed {q_{H_{2}O} \ = \ -5852 \ J}[/tex]
Next, q of liquid nitrogen can be obtained using the calculated q of water:
[tex]q_{N_{2}} \ = -q_{H_{2}O\\\\q_{N_{2}} \ = -(-5852 \ J)\\\boxed {q_{N_{2}} \ = 5852 \ J}[/tex]
The molar heat of vaporization of liquid nitrogen can be calculate using:
ΔH = q/mol N2
[tex]\Delta H _{vap} \ = \frac{5852 \ J}{(\frac{30.0 \ g}{28.0134 g/nol}) } \\\Delta H _{vap} \ = 5464.48056 \ \frac{J}{mol}[/tex]
Since the given has 3 significant figures, the final answer must have 3 significant figures, too.
Hence,
[tex]\boxed {\Delta H _{vap} \ = 5460\ \frac{J}{mol} \ = 5.46 \ x \ 10^3 \ \frac{kJ}{mol}}[/tex]
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Keywords: molar heat of vaporization, calorimetry
