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

The correct answer is 12,48 J.

Step-by-step 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 =
`(m)/(M_(mD) )`

n: Number of moles of Dysprosium

m: Mass of Dysprosium

`M_(mD)`: Molar mass of Dysprosium

n =
`(10 g)/(162,5 (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
`(J)/(mol * C)` * (2600 C - 1400 C)

W=12,48 J

Have a nice day!