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?

Answer 1

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

====> m_iron*S_iron*(100 – t)°C = m_solvent*S_solvent*(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	30	0.786	23.58	58	0.4065
Methanol	30	0.792	23.76	32	0.7425
Ethanol	30	0.789	23.67	46	0.5146
Water	30	1.000	30.00	18	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 Δt = ΔT 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).