Question

How many grams of steam at 100 deg.Celsius would be required to raise the temperature of 43.6 g solid benzene (C6H6) from 5.5 oC to 30.0 oC? Assume that heat is only transfered from the steam (and not liquid water) and that the steam/water and benzene are seperated by a glass wall and do not mix. (The melting point of benzene is 5.5 oC; ΔHfus for benzene is 9.87 kJ/mol; specific heat for benzene is 1.63 J/g ooC; ΔHvap for stream at 100 oC is 40.7 kJ/mol.)

Answer 1

Given that;

Benzene 43.6 g

Temperature change for benzene = 30.0 oC-5.5 oC =24.5 oC

The melting point of benzene is 5.5 oC;

?Hfus for benzene is 9.87 kJ/mol;

specific heat for benzene is 1.63 J/g ooC;

?Hvap for stream at 100 oC is 40.7 kJ/mol

Number of mole of benzene = amount in g/ molar mass

= 43.6 g/ 78.1118 g/ mole

= 0.558 mole benzene

First calculate the amount of heat to melt benzene as follows:

0.558 mole benzene x (9.87 kJ/mol)

= 5.51 kJ

= 5510 J to melt the solid C6H6

Second energy is used to increase temperature of benzene:

= specific heat * mass of benzene * temperature change
(1.63 J/g°C) x (43.6 g C6H6) x (24.5) °C

= 1741.166 J to warm the C6H6 to 30.0°C

5510 J + 1741.166 J = 7251.166 J or 7.25 KJ total required by the C6H6

Heat absorbed by benzene = heat given by steam = mole of steam *?Hvap for stream

Mole of steam = Heat absorbed by benzene / ?Hvap for stream

= 7.25K J/ 40.7 kJ/mol

= 0.1782 mole steam

Amount of steam = number of moles * molar mass

= 0.1782 mole steam * 18.02 g/ mol

= 3.21 g steam at 100 C