The following information is given for n-pentane at 1...

The following information is given for n-pentane at 1 atm: Tb = 36.20°C ?Hvap (36.20°C) = 357.6 J/g Tm = -129.70°C ?Hfus (-129.70°C) = 116.7 J/g Specific heat gas = 1.650 J/g °C Specific heat liquid = 2.280 J/g °C A 50.10 g sample of liquid n-pentane is initially at -49.90°C. If the sample is heated at constant pressure (P = 1 atm), kJ of energy are needed to raise the temperature of the sample to 65.60°C.

Transcript

00:01 Hi, so we are to solve for the amount of heat required in this reaction.

00:08 So first, let us write the temperatures involved.

00:12 We know that the initial temperature is 49 .90 degrees celsius.

00:17 This will be heated up to the final temperature which is 65 .60 degrees celsius, but remember that the boiling point of n -pentane is 36 .20 and then after that we'll have our final temperature here, 65 .60 degrees celsius.

00:38 So from negative 49 .90 degrees celsius to 36 .20 degrees celsius, there will be temperature change and then at 36 .20 degrees celsius, since this is the boiling point, that means there will be phase change from liquid n -pentane to gaseous n -pentane.

00:57 So that's right here, there will be phase change and then from 36 .20 degrees celsius to the final temperature, 65 .60 degrees celsius, that means there will be another change in temperature and we're gonna call the amount of heat required for delta t here, q1.

01:18 This will be q2 and then q3.

01:21 So the total amount of heat required will be equivalent to the sum of q1 plus q2 plus q3.

01:30 So we'll solve them individually.

01:32 We'll start with q1.

01:36 So q1 is the amount of heat required for temperature change, so that means the formula for this will be mass multiplied with a specific heat capacity and then change in temperature.

01:47 Now remember, this is liquid n -pentane.

01:56 So we should take note of the phases of the compound here because the specific heat capacity of liquid n -pentane and then gaseous n -pentane are different.

02:07 Okay, so this is for liquid n -pentane.

02:10 Now we'll plug in the information that we have.

02:14 We have the mass, 50 .10 grams and then the specific heat capacity of liquid n -pentane is 2 .280 joules per gram degree celsius and then the change in temperature will be 36 .20, the final temperature minus negative 49 .90.

02:33 So 36 .20 minus negative 49 .90.

02:40 This is degrees celsius.

02:42 So now we could cancel some units, grams, degrees celsius and the remaining unit will be joules since we're solving for heat.

02:51 So calculating this, we'll have 9835 .0308 joules.

03:00 This is for q1.

03:02 Now for q2, we have phase change.

03:06 This is from liquid to gaseous phase...

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