Question

Question1)

There was a lot of talk among football fans about "deflate-gate" last week (hopefully it's not still going on this week) where the New England Patriots grounds crew seem to have decreased the pressure in the team balls by around 2 PSI from the official value of 12 PSI. (These are Gauge pressures, meaning pressure above 1 atm.) A Physics prof from Boston U. suggested that it may not have been a nefarious plot, but simply a consequence of the fact that it's cold outdoors in Boston and they may have filled the balls indoors. See if this makes sense by calculating the change in pressure (assuming the football had constant volume) between 70 degrees F (nominal inside temperature) and 20 degrees F (a relatively balmy winter-time outside temperature in Boston) from the assumed starting point of 12 PSI.

The Fahrenheit to Celsius conversion is ºC = (ºF – 32)/1.8, there are 14.7 PSI in 1.0 atm, and the volume of a football is 4237 cm³– though you may not even need that last one.

Question 2)

. Imagine you were in the mountains and were desperate for a cup of coffee (or tea, if that's your thing) and the only source of heat you have is a butane cigarette lighter. Could you heat a cup of water at 25 ºC to near boiling with the full capacity of the lighter?

You'll have to make some assumptions about the amount of butane in the lighter and water in your cup, which you should identify; but FYI the density of liquid butane is about 0.6 g/mL, the MW is 58.12 g/mol, and the heat of combustion is -2623 kJ/mol (and of course, the density of water is 1.0 g/mL and the heat capacity is 4.18 J/ºC-g).

Answer 1

A) WE can calculate the pressure by assuming that volume of football is same,

Also if we are not letting the air to come out of the ball then the number of moles of air will also remains the same in the ball

We will use

PV = nRT

If n and v are constant then

Pressure will be directly proportional to temperature

so let intial pressure = P1 = 12 PSI = 0.068 X 12 atm = 0.816 atm

Final pressure P2 = 2 = 0.068 X 2 = 0.136 atm ( we have to check so we will consider it as unknown)

Initial temperature = T1 = 70 degree F = 21.11 oC = 294.26 K

Final temperature = T2 = 20 degree F = 266.48 K

So applying

P1 / P2 = T1 / T2

P2 = P1 X T2 / T1 = 0.816 X 266.48 / 294.26 = 0.738 atm

so the pressure which could reach is 0.738

So the decrease in pressure is due to some other reason as well

2) The combustion equation will be

C4H10 + 13/2 O2 --> 4CO2 + 5H2O   which will give energy = -2623. KJ / moles

so heat evolved by 58.12 grams of butane = 2623. KJ

Heat evolved by 1 gram = 2623 / 58.12 = 45.13 KJ

Let there is 10mL of butan in lighter

Then gram of butane = density X volume = 0.6 X 10mL = 6 grams

So heat evolved = 45.13 X 6 KJ = 270.78 KJ

If we wish to heat the water to 100 ⁰C from 25_⁰C ,

The amount of water which could be heated can be caluclated as

Heat required by water = Mass X Heat capacity X change in temperature = Heat given by buring of butane

270.78 X 1000 Joules = Mass X 4.18 X (100-25)

Mass = 863.73 grams or 863.73 mL

So yes we can heat this much amount of water .

However for converting water to steam we need more energy , that will be heat of evaporation of water