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theodore100
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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.