Chemistry help needed by 5:30!!!

Question

The melting point of substance X is 145°C and the boiling point of X is 545°C. The specific heat of x As a solid is 1.24 J per gram degrees Celsius, 0.960 J per gram Celsius as a liquid, and 0.565 J per gram degrees Celsius as a gas. The heat of fusion of X is 232 J per gram and the heat of vaporization of X is 985 J/g.

• How many joules of energy are lost to cool 0.784 g of X from 811°C to 331°C?

• show values of q1,q2,q3

•what is q total?

Answer 1

Well, this answer missed the deadline, but is included for future reference.

MP = 145º

C_Solid =1.24 J/gdeg

BP = 545º

C_liq = 0.960 J/gdeg

C_gas = 0.565 J/gdeg

∆H_fusion = 232 J/g

∆H_vap = 985 J/g

mass = 0.784 g

Initial temp = 811º

Final temp = 331º

(1). To cool 0784 g from 811º(gas) to 545º(gas) = q1 = mC∆T = (0.784g)(0.565 J/gdeg)(266deg) = 117.8 J

(2). To condense 0.784 g into liquid at 545º = q2 = m∆H_vap = (0.784 g)(985 J/g) = 772.2 J

(3). To cool 0.784 g as a liquid from 545º to 331º = q3 = mC∆T = (0.784g)(0.960J/gdeg)(214deg) = 161 J

Total q = q1 + q2 + q3 = 1051 = 1050 J (to 3 sig. figs.)