Simple heat transfer problem and temperature change

[theodore100]

Homework Statement

What is the expected temperature change during one second of a polypropylene solid with mass 0.1 g at 105 degC to air at 20 degC

Homework Equations

heat transfer: Q= c m (T1-T0)
where Q = amount of heat
c = specific heat capacity = 1.8 J/g/K
m = mass
T1-T0 = change in temperature

Newton's law thermal cooling:
Q = -h.A.dT(t)

Q = heat lost
h = heat transfer coefficient for polypropylene = 0.2
A = area, small object = approx. 0.5 cm^2
Temperature differential, constant at 105-20

The Attempt at a Solution

T1-T0 = Q / c m = Q / 1.8 * 0.1

Therefore:

T1-T0 = (-0.2 * 5*10^-5 * 85) / (1.8 * 0.1) = 4.7 * 10^-3 degC

i.e. a very very small temperature drop in one second. Does this look right?

Thanks.

 

[gneill]

Hi theodore100, Welcome to Physics Forums.

Be careful with your unit conversions. 1 gram is not 0.1kg.

 

[theodore100]

Hi and thanks for the welcome.
I have got my units a bit mixed up, but the 0.1 is for the c * m calculation, c being in units of J/g/K. I will try it again with everything in Kg.

 

[theodore100]

I get the same answer:
c in J/kg/K = 1800

T1-T0 = (-0.2 * 5*10^-5 * 85) / (1800 * 1*10^-4) = -0.0047

 

[theodore100]

I'm guessing that the area has to be in m and 0.5 cm^2 is 5*10^-5 m^2 ?

 

[gneill]

Okay, provided that your units are in order, then your result looks good.

This presumes that the constant h really is the heat transfer coefficient for polypropylene to air (I've seen similar values for heat conductivity for polypropylene and it's curious that they would be the same value). Were you given units for h?

 

[theodore100]

No, I wasn't given h, I googled it.. so yes, it might be wrong.

 

[theodore100]

ok, this is interesting.. it might be more like 24 W/m^2 K
so the answer could be as high as 0.564 which sounds more realistic. I think I have the right approach, which is the main thing. Thanks.

 

[gneill]

Okay.

I should also note that heat loss by radiation is being ignored. Emissivity of PP is pretty high, around 0.97 if I recall correctly. You might check to see which mode of heat loss would dominate.