Use the following information to answer question 19 and 20. The enthalpy of comb
ID: 949638 • Letter: U
Question
Use the following information to answer question 19 and 20. The enthalpy of combustion of octane is 5074.9 kJ/mol and its density is 703 kg/m^3. On a mild winter day a room of volume 40 m^3 needs to be warmed up from 10 to 20DegreeC? (c_air - 0.718 J/g K; d_air - 1.225 g/L). How much heat is needed? 27 kJ 35 lK 49 kJ 57 kJ None of the above How much octane needs to be burned? 0.845 L 1.048 L 1.124 L 1.648 L None of the above Which of the following algebraic expressions represents the change in enthalpy between solid water (ice) at -20DegreeC and liquid water at + 25DegreeC? nDelta H^H_2O^-20DegreeC(s) Right H_2O^+25DegreeC(t) = mc_ice (0DegreeC - ( - 20DegreeC)) + nDelta H_fus^H_2O,ODegreeC + mc_water(+25DegreeC + 0DegreeC) nDeltaH^H_2O^-20DegreeC(s) Right arrow H_2O^+25DegreeC(t) = nDelta H_fus^H_2O,0DegreeC + mc_water (+25DegreeC - (-20Degree C)) nDeltaH^H_2O^-20DegreeC(s) Right arrow H_2O_(t)^25DegreeC = nDeltaH_fus^H_2O,0DegreeC Any of the above may be used, as it is a state function and it is therefore path independent None of the above is correctExplanation / Answer
19: Given the volume of air, V = 40 m3 = 40 m3 x (1000 L / 1m3) = 40000 L
Mass of air inside the room, m = Vxd = 40000 L x (1.225 g /L) = 49000 g
Heat required to increase temperature of air from 10 C to 20 C,
Q = mxc(air)xdT = 49000g x 0.718 J/(g.K) x 10 K = 351820 J = 352 KJ (answer)
Hence (E). non of the given options are correct.
(20); given the heat of combustion of octane = 5074.9 KJ/mol
Molecular mass of octane = 114.23 g/mol
i.e 114.23 g of octane gives 5074.9 KJ heat.
Hence 352 KJ of heat that will be produced by the amount of octane, m = (114.23 g / 5074.9 KJ) x 352 KJ
= 7.923 g
Density of octane, d = 703 Kg/m3 = 703 g/L
Hence volume of octane required, V = m/d = 7.923 g / 703 g/L = 0.0113 L (answer)
Hence (E). non of the given options are correct.
(21): When ice at - 20 C is converted to water at 25 C, it has to pass through 3 different steps, which are from ice at -20 C to ice at 0 C, from ice at 0 C to water at 0 C, from water at 0 C to water at 25 C.
Hence option (A) is correct.