College Chemistry question, I cant seem to work out the problem

Question

Methylene chloride has a melting point of-94.8°C, a boiling point of 40.0°C, specific heat of 1.21 J/g*K for the solid and liquid, and 0.608 J/g*K for the gas. The heats of fushion and vaporization are ◇Hfus of 73.40 J/g and ◇Hvap of 334.35 J/g respectively. Determine how many joules of energy are needed to heat 20g of methylene chloride from -171.4°F to 333K.

Answer 1

Easiest to do these problems in steps, noting that methylene chloride (mc) freezes/melts at -94.8ºC and boils/vaporizes at 40.0ºC. Also, note different units of temperature that are given in this problem (sneaky).

Let us first convert these so they are all the same.

T_(°C) = (T_(°F) - 32) × 5/9

ºC = (-171.4F - 32) x 5/9 = -113ºC = initial temperature

333K - 273 = 60ºC = final temperature

Then we use q = mC∆T when there is no phase change, and q = m∆H when there is a phase change. Note that since we are using ∆T (i.e. change in temp) the units of K or C don't matter.

Step 1. heat needed to raise temp of 20 g mc from -113º to -94.8º = q = (m)(C)(∆T) = (20g)(1.21J/g/º)(18.2º) = 440 J

Step 2. heat needed to melt 20 g mc at -94.8º = q = m∆Hf = (20g)(73.40J/g) = 1468 J

Step 3. heat needed to raise temp of 20 g mc from -94,8º to 40º = q = (m)(C)(∆T) = (20g)(1.21J/g/º)(134.8º) = 3262 J

Step 4. heat needed to vaporize 20 g mc at 40º = q = m∆Hvap = (20g)(334.35 J/g) = 6687 J

Step 5. heat needed to raise temp of 20 g mc from 40º to 60ºC = q = (m)(C)(∆T) = (20g)(0.608J/g/º)(20º) = 243 J

Add up all the joules to get the total amount of energy needed:

440J + 1468J + 3262J + 6687J + 243J = 12,100 J