you have a 4 10g cubes of iron at 100°C and you place them into 30ml of acetone,

ID: 560911 • Letter: Y

Question

you have a 4 10g cubes of iron at 100°C and you place them into 30ml of acetone, methanol, ethanol, and water respectively, what will the temperature of the solvents be after the placement of the iron? (they start at 20°C) Specific heat capacity of Iron is .450 J/g C, the specific heat capacity of acetone is 125.5 J/(mol K), the specific heat capacity of methanol is 79.5 J/(mol K), the specific heat capacity of ethanol is 112.4 J/(mol K), the specific heat capacity of water is 75.26 J/(mol K). Will any of the solvents reach the boiling point?

Explanation / Answer

Let the temperature attained by the hot substance (iron) and the cold substance (solvent) be t°C. We shall employ the principle or thermochemistry:

Heat lost by the hot substance = Heat gained by the cold substance

====> miron*Siron*(100 – t)°C = msolvent*Ssolvent*(t – 20)°C

where m denotes mass and S denotes specific heat capacity.

List down the densities and the molar masses of the solvents and calculate the mole(s) of wach solvent.

Solvent

Volume of solvent (mL)

Density of solvent (g/mL)

Mass of solvent (g) = volume*density

Molar mass of solvent (g/mol)

Mole(s) of solvent = mass/molar mass

Acetone

0.786

23.58

0.4065

Methanol

0.792

23.76

0.7425

Ethanol

0.789

23.67

0.5146

Water

1.000

30.00

1.6667

Use the principle of thermochemistry and substitute values to find t.

Acetone:

(10.0 g)*(0.450 J/g.°C)*(100 – t) = (0.4065 mole)*(125.5 J/mol.K)*(t – 20) K (note that t = T since both Celcius and Kelvin scales have 100 graduations between the LFP and UFPs]

===> 4.5*(100 – t) = 51.01575*(t – 20)

===> 450 – 4.5t = 51.01575t – 1020.315

===> 450 + 1020.315 = 55.51575t

===> t = 26.4846 26.49

The final temperature reached is 26.49°C (ans).

Methanol:

(10.0 g)*(0.450 J/g.°C)*(100 – t) = (0.7425 mole)*(79.5 J/mol.K)*(t – 20) K

===> 4.5*(100 – t) = 59.02875*(t – 20)

===> 450 – 4.5t = 59.02875t – 1180.575

===> 450 + 1180.575 = 63.52875t

===> t = 25.6667 25.67

The final temperature reached is 25.67°C (ans).

Ethanol:

(10.0 g)*(0.450 J/g.°C)*(100 – t) = (0.5146 mole)*(112.4 J/mol.K)*(t – 20) K

===> 4.5*(100 – t) = 57.84104*(t – 20)

===> 450 – 4.5t = 57.84104t – 1156.8208

===> 450 + 1156.8208 = 62.34104t

===> t = 25.7747 25.78

The final temperature reached is 25.78°C (ans).

Water:

(10.0 g)*(0.450 J/g.°C)*(100 – t) = (1.6667 mole)*(75.26 J/mol.K)*(t – 20) K

===> 4.5*(100 – t) = 125.4358*(t – 20)

===> 450 – 4.5t = 125.4358t – 2508.716

===> 450 + 2508.716 = 129.9358t

===> t = 22.7706 22.77

The final temperature reached is 22.70°C (ans).