An electronic device dissipating 30 W has a mass of 20 g and a specific heat of

Question

An electronic device dissipating 30 W has a mass of 20 g and a specific heat of 850 J/kg middot degree C. The device is lightly used, and it is on for 5 min and then off for several hours, during which it cools to the ambient temperature of 25 degree C. Determine the highest possible temperature of the device at the end of the 5-min operating period. What would your answer be if the device were attached to a 0.2-kg aluminum heat sink? Assume the device and the heat sink to be nearly isothermal. Reconsider Prob. 5-92. Using EES (or other) software, investigate the effect of the mass of the heat sink on the maximum device temperature. Let the mass of heat sink vary' from 0 to 1 kg. Plot the maximum temperature against the mass of heat sink, and discuss the results. An ordinary egg can be approximated as a 5.5-cm-diameter sphere. The egg is initially at a uniform temperature of 8 degree C and is dropped into boiling water at 97 degree C. Taking the properties of the egg to be rho = 1020 kg/m^3 and c_p = 3.32 kJ/kg middot degree C, determine how much heat is transferred to the egg by the time the average temperature of the egg rises to 80 degree C. In a production facility, 1.2-in-thick 2-ft times 2-ft square brass plates (rho = 532.5 lbm/ft^3 and C_p = 0.091 Btu/lbm middot degree F) that are initially at a uniform temperature of 75 degree F are heated by passing them through an oven at 1300 degree F at a rate of 300 per minute. If the plates remain in the oven until their average temperature rises to 1000 degree F, determine the rate of heat transfer to the plates in the furnace.

Explanation / Answer

Given

the egg diameter d = 5.5 cm , radius r = 2.75 cm = 0.0275 m,

   volume is v = 4/3Pi*r^3= (4/3)pi*0.0257^3 m3 = 71.103*10^-6 m3

   density is rho = 1020 kg/m3 ,

mass of egg is m = rho*v = 1020*71.103*10^-6 kg = 0.07252506 kg

heat capacity of egg Cp = 3320 J/kg 0C

tmeperature difference is DT = 72 0C

now heat absorbed by egg is Q = m*Cp*DT
           Q = 0.07252506*3320*72 J

           q = 17336.390 J

and

heat capacity of water is 4186 J/kg 0C, mass of water of 1 lt = 1kg

Temperature difference of water is DT = 97-80 = 17

   17336.390 J = 1*4186(370-T)

   T = 93 0C

temperature drop is 7 0C from 97 0C to 93 0C

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