A gas stream containing 45wt% butane and 55wt% pentane is subjected to heating from 180K to 260K at constant pressure. calculate the required heat input per kg of this mixture?

neglecting changes in kinetic and potential energies.

Jun 11, 2018

The required heat input is 184 kJ/kg.

Explanation:

Assume that we have 1 kg of the mixture.

Then we have 450 g of butane and 550 g of pentane.

Step 1. Calculate the heat input for the butane

Butane is a liquid between 180 K and 260 K.

The formula for the quantity of heat $q$ transferred is

color(blue)(bar(ul(|color(white)(a/a)q = mC_text(s)ΔTcolor(white)(a/a)|)))" "

where

$m \textcolor{w h i t e}{l l} =$ the mass of the object
${C}_{\textrm{s}} \textcolor{w h i t e}{l} =$ its specific heat capacity
ΔT = its change in temperature

In this problem,

$m \textcolor{w h i t e}{m} = \text{450 g}$
${C}_{\textrm{s}} \textcolor{w h i t e}{l l} = \text{132.42 J·K"^"-1""mol"^"-1}$
ΔT = T_text(f) - T_text(i) = "260 K - 180 K = 80 K"

The first task is to convert the mass of the butane to moles.

n = 450 color(red)(cancel(color(black)("g"))) × "1 mol"/(58.12 color(red)(cancel(color(black)("g")))) = "7.743 mol"

Now we can determine the quantity of heat needed.

q_text((butane) = 7.743 color(red)(cancel(color(black)("mol"))) × "132.42 J"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 80 color(red)(cancel(color(black)("K"))) = 8.20 × 10^4color(white)(l)"J"

Step 2. Calculate the heat input for the pentane

Pentane is a liquid between 180 K and 260 K.

Here,

$m \textcolor{w h i t e}{m} = \text{550 g}$
${C}_{\textrm{s}} \textcolor{w h i t e}{l l} = \text{167.19 J·K"^"-1""mol"^"-1}$

We must convert the mass of the pentane to moles.

n = 550 color(red)(cancel(color(black)("g"))) × "1 mol"/(72.15 color(red)(cancel(color(black)("g")))) = "7.623 mol"

Now we can determine the quantity of heat needed.

q_text(pentane) = 7.623 color(red)(cancel(color(black)("mol"))) × "167.19 J"·color(red)(cancel(color(black)("K"^"-1""mol"^"-1"))) × 80 color(red)(cancel(color(black)("K"))) = 1.02 × 10^5color(white)(l)"J"

Step 3. Calculate the total heat required

q_text(total) = q_text(butane) + q_text(pentane) = (8.20 × 10^4 + 1.02 × 10^5)color(white)(l)"J" = 1.84 × 10^5 color(white)(l)"J" = "184 kJ"