   Chapter 11, Problem 70AP

Chapter
Section
Textbook Problem

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 (ill the need. The device consists of a foam-insulated box containing several packets of a waxy material that melts at 37.0°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 0.490 m2, thickness 9.50 cm, and conductivity 0.012 0 W/m · °C Assume the exterior temperature is 23.0°C for 12.0 h and 16.0°C for 12.0 h. (a) What mass of the waxy material is required to conduct the bacteriological test? (b) Explain why your calculation can be done without knowing the mass of the test samples or of the insulation.

(a)

To determine
The mass of waxy material required to conduct the bacteriological test.

Explanation

Given Info: Melting point of waxy material is 37.0°C , Latent heat of fusion of material is 205kJ/kg , surface area of the insulation panel is 0.490m2 , thickness 9.50 cm, conductivity of the material 0.0120W/m°C , exterior temperature is 23.0°C for 12 h, and interior temperature is 16.0°C for 12 h.

Formula to find rate at which energy is being transferred to the sample is,

P=kA(ThTc)L

• P is rate at which energy is being transferred,
• k is thermal coefficient of wax,
• A is the insulation panel,
• Th is the interior temperature,
• Tc is the exterior temperature,
• L is the thickness of the insulation panel,

Formula to calculate the heat energy lost by conduction is,

Q=P1t+P2t

• Q is the heat energy lost by conduction through insulation,
• P1 is the rate of heat lost by conduction for the first 12 h,
• P2 is the rate of heat lost by conduction for the second 12 h,
• t is the time of heat transfer,

Use kA(ThTc1)L for P1 and kA(ThTc2)L for P2 in the above equation to rewrite Q.

Q=(kA(ThTc1)L)t+(kA(ThTc2)L)t=kALt[(ThTc1)+(ThTc2)]

Substitute 0.0120J/s.m°C for k, and 0.490m2 for A, 9

(b)

To determine
Why the calculation can be done without knowing the mass of the test samples or of the insulation.

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