Thermodynamics: What mass of the waxy material is required to conduct the test?

AI Thread Summary
The discussion revolves around calculating the mass of a waxy material needed for a low-cost incubator designed for bacteriological testing at a constant temperature of 37°C. The user initially misapplied the temperature difference in their calculations, mistakenly using the exterior temperatures instead of the required incubation temperature. After clarification, it was established that the correct temperature difference should be between the incubation temperature (37°C) and the lower exterior temperature. The user adjusted their calculations accordingly and confirmed they arrived at the correct solution. The conversation highlights the importance of understanding thermodynamic principles in practical applications.
HermanC
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Homework Statement



For bacteriological testing of water supplies and in medical clinics, samples must routinely be incubated for 24 h at 37°C. A standard constant temperature bath with electric heating and thermostatic control is not suitable in developing nations without continuously operating electric power lines. Peace Corps volunteer and MIT engineer Amy Smith invented a low cost, low maintenance incubator to fill the need. The device consists of a foam-insulated box containing several packets of a waxy material that melts at 37°C, interspersed among tubes, dishes, or bottles containing the test samples and growth medium (food for bacteria). Outside the box, the waxy material is first melted by a stove or solar energy collector. Then it is put into the box to keep the test samples warm as it solidifies. The heat of fusion of the phase-change material is 205 kJ/kg. Model the insulation as a panel with surface area 0.530 m2, thickness 9.40 cm, and conductivity 0.0120 W/m·°C. Assume the exterior temperature is 24.5°C for 12.0 h and 15.5°C for 12.0 h. (a) What mass of the waxy material is required to conduct the bacteriological test?

Homework Equations



P = kA(Th-Tc) / L
P = Q / Δt

The Attempt at a Solution



This is what I tried doing:
P = kA(Th-Tc) / L = (0.0120 W/m C)(0.530 m2)(24.5 C - 15.5 C) / 0.0940 m = 0.609 W
P = Q / Δt = mL / Δt --> m = PΔt / L = (0.609W)(24 h * 3600 s/h) / (205 kJ/kg * 1000 J/kJ) = 0.257 kg
What am I doing wrong? Thanks in advance.
 
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Hi HermanC, welcome to PF. The first thing that stands out in your solution is the "24.5 C - 15.5 C" part. Why are you using the difference between these two temperatures?
 
Thank you Mapes.

I'm not sure what the last sentence of the problem means, but I assumed 24.5 C - 15.5 C was the change in temperature and 12.0 h + 12.0 h was the change in time.
 
But Th-Tc is the difference in temperature across a distance. What's Th in this problem?
 
37 C.

I got it right. Thanks.
 
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