Respuesta :
Answer:
The quantity of heat necessary to change the temperature of 3.00 mol of the substance from 27°C to 227°C is 19.668 KJ
Explanation:
From the question, The empirical equation is
C=29.5J/(mol⋅K)+(8.20×10−3J/(mol⋅K2))T
[tex]C=29.5J/(mol.K)+(8.20\times 10^{-3} J/(mol.K^{2} ))T[/tex]
Now, to determine the heat necessary to change the temperature of 3.00 mol of this substance from 27∘C to 227∘C, that is ΔQ
From, dQ=nCdT
Integrating both sides, we get
[tex]\int\limits^ {Q_{2}} _{Q_{1}} {dQ} \, = \int\limits^ {T_{2} }_{T_{1} } nC \, dT[/tex]
[tex]{Q_{2}} -{Q_{1}} = \int\limits^ {T_{2} }_{T_{1} } n[ {29.5J/(mol.K)+(8.20\times 10^{-3} J/(mol.K^{2} ))T} \, ]dT[/tex]
[tex]\Delta Q = \int\limits^ {T_{2} }_{T_{1} } 3.00mol[ {29.5J/(mol.K)+(8.20\times 10^{-3} J/(mol.K^{2} ))T} \, ]dT[/tex]
[tex]\Delta Q = \int\limits^ {T_{2} }_{T_{1} } [ {88.5J/K+ (24.6\times 10^{-3} J/K^{2} )T} \, ]dT[/tex]
[tex]\Delta Q = \int\limits^ {T_{2} }_{T_{1} } {88.5J/KdT + \int\limits^ {T_{2} }_{T_{1} } 24.6\times 10^{-3} J/K^{2} T} dT[/tex]
[tex]\Delta Q =(88.5J/K )\int\limits^ {T_{2} }_{T_{1} } dT + (24.6\times 10^{-3} J/K^{2})\int\limits^ {T_{2} }_{T_{1} } T} dT[/tex]
(NOTE: [tex]\int {dx} = x[/tex] and [tex]\int\ {x} \, dx = \frac{x^{2} }{2}[/tex] )
Hence, we get
[tex]\Delta Q =(88.5J/K )({T_{2} }-{T_{1} )+ (24.6\times 10^{-3} J/K^{2}) (\frac{T_{2}^{2} }{2} - \frac{T_{1}^{2} }{2} )[/tex]
From the question, [tex]T_{1}[/tex] = 27 °C = (27+273) K = 300K
Also, [tex]T_{2}[/tex] = 227 °C = (227+273) K = 500K
Then,
[tex]\Delta Q =(88.5J/K )(500K - 300K )+ (24.6\times 10^{-3} J/K^{2}) (\frac{(500K)^{2} }{2} - \frac{(300K)^{2} }{2} )[/tex]
[tex]\Delta Q =(88.5J/K )(200K )+ (24.6\times 10^{-3} J/K^{2}) (80\times 10^{3} K^{2} )[/tex]
[tex]\Delta Q =17700J +1968J \\[/tex]
[tex]\Delta Q =19668J[/tex]
[tex]\Delta Q =19.668KJ[/tex]
Hence, the quantity of heat necessary to change the temperature of 3.00 mol of the substance from 27°C to 227°C is 19.668 KJ
(NOTE: KJ means Kilo Joules)
