the following information is given for n pentane at 1 atm 3620c 3620c jg 12970c 12970c jg specif

Step 1: Understand the problem. We need to calculate the total energy required to heat a 28.00 g

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

The following information is given for n-pentane at 1 atm: T1 = 36.20°C (T1) = J/g T2 = –129.70°C (T2) = J/g Specific heat gas = 1.650 J/g°C Specific heat liquid = 2.280 J/g°C A 28.00 g sample of liquid n-pentane is initially at –91.80°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 64.80°C.

Transcript

00:01 In this problem, we have n -pentane at one atmosphere.

00:05 We are also given the corresponding boiling points and melting points of our n -pentane.

00:11 Additionally, we are given the latent heat of vaporization and the latent heat of fusion.

00:18 We also have the corresponding specific heat values for our n -pentine in gas form and our n -pentine in liquid form.

00:28 So now, suppose that we have a certain sample of and pentane at 28 grams.

00:39 And we want to see how much heat is required if we have a starting temperature of negative 91.

00:50 Negative 91 .8 degrees celsius.

00:55 And we heat this sample of pentane to a final temperature of 64 .8 degrees celsius.

01:07 So how much heat will this take? so let's write in our equation for heat.

01:13 Q is equal to mass of the sample times specific heat, times the change in temperature, plus our mass times the respective latent heat.

01:26 And what we're going to do is break down our calculations for our total q in segments of q1, q2, and q3.

01:39 So in our q1 calculation, we of course have the mass of our sample.

01:49 And initially, if we're starting at a temperature of negative 91 .8, our pentane will be in its liquid state.

01:59 So we're going to write the specific key value of our, i guess, our n -pentine in liquid form.

02:07 So c, lick.

02:08 And our delta t will be equal to until the, i guess until our pentane boils minus our initial temperature.

02:24 So essentially we're going to have a liquid sample up until this boiling point.

02:31 And so doing this calculation, we get 28 times our specific heat in liquid form times our temperature.

02:47 Change.

02:47 So again, if we're going, i guess going up to our boiling point, our boiling point is 36 .2 degrees celsius.

02:56 So i'll be 36 .2 degrees celsius...

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