Question

Dysprosium was discovered in 1886. It's freezing point is 1400C and its boiling point is 2600C. If it's specific heat is 0.1733J/g.C, how many joules are required to heat 10.0g from its freezing point to its boiling point?

Answer 1

Answer:

The correct answer is 12,48 J.

Explanation:

According to the first law of thermodynamics, the variation of the internal energy is equal to the heat minus the work.

ΔU = Q - W

Being:

Q=n * C * ($T_{f}$-$T_{0}$)

Q: Heat

n: number of moles

C: specific heat

$T_{f}$: Final temperature

$T_{0}$: Initial temperature

As in this situation there is no variation of the internal energy, the initial equation remains:

W=Q

First, we calculate the number of moles of Dysprosium present in a 10g sample.

n = $\frac{m}{M_{mD} }$

n: Number of moles of Dysprosium

m: Mass of Dysprosium

$M_{mD}$: Molar mass of Dysprosium

n = $\frac{10 g}{162,5 \frac{g}{mol} }$

n=0,06 mol

Now, replacing the values given in the problem in the equation of Q, we calculate W as:

W = n * C * ($T_{f}$-$T_{0}$)

W=0,06 mol * 0,1733 $\frac{J}{mol * C}$ * (2600 C - 1400 C)

W=12,48 J

Have a nice day!