Saturated acetone vapor enters a heat exchanger at a rate of 2.50 L/s at 1 atm,

Question

Saturated acetone vapor enters a heat exchanger at a rate of 2.50 L/s at 1 atm, and emerges from the heat exchanger as a saturated liquid at the same pressure. Water at 1 atm enters the heat exchanger at 18.0 degree C and at a rate of 9.32 mol/s. Assume that the heat exchanger operates adiabatically with respect to its surroundings. Calculate the temperature of exiting water by following the steps below. At what temperature does acetone enter the heat exchanger? What is the heat of vaporization of acetone at this temperature? Use the ideal gas law to calculate the moles of acetone entering the heat exchanger. Assume that the heat capacity of water for the temperature range in the heat exchanger is constant at 7.54 times 10^-2 kJ/(mol middot degree C). What is the temperature of water leaving the heat exchanger?

Explanation / Answer

Boiling point of acetone at 1atm= 56 deg.c

Saturation temperature of acetone = 56 deg.c

P= 1atm T= 56 deg.c= 56+273= 329K, R= 0.0821 L.atm/mole.K

V= 2.5 L/s, moles of acetone vapor n= PV/RT = 1*2.5/(0.0821*329)=0.093 mol/s

Heat of vaporization of acetone = 32 Kj/mol

Acetone leaves as saturated liquid at 56 deg.c. Hecne acetone looses latent heat of vaporization

Heat lost by acetone = 0.093 mol/s* 32Kj/mol =2.976 Kj= 2.976*1000 Joules/se = 2976 joules/s

This heat needs to be removed for this adiabatic system.

Hence 2976= molar flow rate of water* specific heat of water* (T4-T3)

2976= 9.32 mol/s * 7.54*10-2*1000 *(T4-18)

T4=22.23 deg.c

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