Question

The molar heat capacity of a certain substance varies with temperature according to the empirical equation C=29.5J/(mol⋅K)+(8.20×10−3J/(mol⋅K2))T How much heat is necessary to change the temperature of 3.00 mol of this substance from 27∘C to 227∘C ? (Hint: Use the equation dQ=nCdT and integrate.)

Answer 1

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

Step-by-step explanation:

From the question, The empirical equation is

C=29.5J/(mol⋅K)+(8.20×10−3J/(mol⋅K2))T

`C=29.5J/(mol.K)+(8.20* 10^(-3) J/(mol.K^(2) ))T`

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

`\int\limits^ {Q_(2)} _{Q_(1)} {dQ} \, = \int\limits^ {T_(2) }_{T_(1) } nC \, dT`

`{Q_(2)} -{Q_(1)} = \int\limits^ {T_(2) }_{T_(1) } n[ {29.5J/(mol.K)+(8.20* 10^(-3) J/(mol.K^(2) ))T} \, ]dT`

`\Delta Q = \int\limits^ {T_(2) }_{T_(1) } 3.00mol[ {29.5J/(mol.K)+(8.20* 10^(-3) J/(mol.K^(2) ))T} \, ]dT`

`\Delta Q = \int\limits^ {T_(2) }_{T_(1) } [ {88.5J/K+ (24.6* 10^(-3) J/K^(2) )T} \, ]dT`

`\Delta Q = \int\limits^ {T_(2) }_{T_(1) } {88.5J/KdT + \int\limits^ {T_(2) }_{T_(1) } 24.6* 10^(-3) J/K^(2) T} dT`

`\Delta Q =(88.5J/K )\int\limits^ {T_(2) }_{T_(1) } dT + (24.6* 10^(-3) J/K^(2))\int\limits^ {T_(2) }_{T_(1) } T} dT`

(NOTE:
`\int {dx} = x` and
`\int\ {x} \, dx = (x^(2) )/(2)` )

Hence, we get

`\Delta Q =(88.5J/K )({T_(2) }-{T_(1) )+ (24.6* 10^(-3) J/K^(2)) ((T_(2)^(2) )/(2) - (T_(1)^(2) )/(2) )`

From the question,
`T_(1)` = 27 °C = (27+273) K = 300K

Also,
`T_(2)` = 227 °C = (227+273) K = 500K

Then,

`\Delta Q =(88.5J/K )(500K - 300K )+ (24.6* 10^(-3) J/K^(2)) (((500K)^(2) )/(2) - ((300K)^(2) )/(2) )`

`\Delta Q =(88.5J/K )(200K )+ (24.6* 10^(-3) J/K^(2)) (80* 10^(3) K^(2) )`

`\Delta Q =17700J +1968J \\`

`\Delta Q =19668J`

`\Delta Q =19.668KJ`

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)