The fluorocarbon compound c2cl3f3 has a normal boiling point of 47.6 *c. the specific heats of c2cl3f3 liquid and gas are 0.91 j/g*k (l) and 0.67 j/g*k (g), respectively. the heat of vaporization for the compound is 27.49 kj/mol. calculate the heat required in kilojoules to convert 38.0 g of c2cl3f3 from a liquid at 12.30*c to a gas at 93.15*c.

7916.59 J of heat is required

Explanation:

the total heat required is

total heat = sensible heat to heat the liquid until boiling + heat to vaporise all the compound and take it to the gas state + heat required to heat the gas until the final temperature

Q = Q1+ Q2+ Q3

where

Q1 = m * cL * ΔTL

Q2 = m * Lv

Q3= m * cG * ΔTG

cL and cG are the specific heats of the liquid and gas respectively . ΔTL and ΔTG are the temperature changes for liquid and gas. Lv is the molar latent heat of vaporisation.

the mass based latent heat of vaporisation is

Lv m = Lv / M = 27.49 kJ/mol / 187.5 gr/mol =0.1466 kJ/g , where M= molecular weight of C2Cl3F3 ( calculated by summing all the molecular weights of its elements)

therefore

Q1 = m * cL * ΔTL = 38 g * 0.91 J/gK * ( 47.6°C - 12.3°C) = 1186.09 J

Q2 = m * Lv m= 38 g * 0.1466 kJ/g * 1000J/kJ = 5570.8 J

Q3 = m * cG * ΔTG = 38 g * 0.67 J/gK * (93.15°C - 47.6°C ) =  1159.70 J

thus

Q = Q1+Q2+Q3 = 1186.09 J + 5570.8 J + 1159.70 J = 7916.59 J

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

not too sure thinking all accelerations are pointed down