# Simple heat transfer problem and temperature change

1. Oct 24, 2013

### theodore100

1. The problem statement, all variables and given/known data
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

2. Relevant 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

3. 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.

2. Oct 24, 2013

### Staff: Mentor

Hi theodore100, Welcome to Physics Forums.

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

3. Oct 24, 2013

### 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.

4. Oct 24, 2013

### theodore100

c in J/kg/K = 1800

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

5. Oct 24, 2013

### theodore100

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

6. Oct 24, 2013

### Staff: Mentor

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?

7. Oct 24, 2013

### theodore100

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

8. Oct 24, 2013

### 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.

9. Oct 24, 2013

### Staff: Mentor

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.