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?

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 * (-)

Q: Heat

n: number of moles

C: specific heat

: Final temperature

: 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 =

n: Number of moles of Dysprosium

m: Mass of Dysprosium

: Molar mass of Dysprosium

n =

n=0,06 mol

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

W = n * C * (-)

W=0,06 mol * 0,1733 * (2600 C - 1400 C)

W=12,48 J

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i just learnt this recently but i need to look through my notes again