Answer:
4.86 x 10⁴ J are released
Explanation:
We need to calculate the heat released when diethyl ether at 53 ºC is condensed and cooled to a temperature of 10 ªC.
So the process is divided in three steps:
1.- Cool the gas to the normal boiling point
2.-Phase change from gas to liquid
3.- Cool the liquid from 34.5 ºC to 10 ºC
ΔH for step 1:
ΔH₁ = m x csp x ΔT
where m is the mass, csp is the specific heat and ΔT is the change in temperature ( Tfinal - Tinicial ). Here we will be using the specific heat for (CH3CH2)2O(g), 2.35 J/g·ºC.
= 100 g x 2.35 J/g·ºC x ( 34.5 -53 )ºC = -4.35 x 10³ J
ΔH for step 2:
ΔH₂ condensation = mC heat of vaporization
= 100.0 g x 351 J/g = -3.51 x 10⁴ J
ΔH₃ for step 3:
ΔH₃ = m x csp x ΔT
Here we will the specific heat for the liquid since now we have the ether in the liquid phase.
ΔH = 100 g x 3.74 J/g·ºC x ( 10.0 - 34.5 )ºC = - 9.16 x 10³ J
The total energy change is:
ΔHtotal = ΔH₁ + ΔH₂ + ΔH₃ = -4.35 x 10³ J + ( -3.51 x 10⁴ J )= + (- 9.16 x 10³ J)
ΔHtotal =-4.86 x 10⁴ J
The sign is negative since heat is released.
