As a coffee enthusiast you wish to keep your coffee as hot as possible for as lo

Question

As a coffee enthusiast you wish to keep your coffee as hot as possible for as long as possible. You wish to find the rate of heat loss of a typical coffee mug. Assume the mug to be a cylinder which has a lid made out of the same material as the rest of the cylinder, and the thickness of the walls is 1.00 cm. If the coffee in the mug is 81.3 °C and the room temperature is 20.8 °C how much more heat does the mug lose through conduction than from radiated energy. See the hint below for important constants.

HINT: Assume that the thermal conductivity of all sides of the mug is 0.873 W/m·C, that the mug has an emissivity of 0.760 and the Stefan-Boltzmann Constant is 5.67 × 10-8 W/m2·K4. Remember to properly convert all units and to include all surfaces of the mug.

Explanation / Answer

let,

thick ness of the walls of the cylinder, l=1cm

temeperature T1=81.3 oC =354.3 K

temeperature T2=20.8 oC =293.8 K

e is emissivity e=0.76

thermal conductivity k=0.873

stefan constant, sigma=5.67*10^-8 W*m^2/K4

a)

rate of heat energy per area through conduction process is,

I1=P/A=k*(T1-T2)/l

I1=0.873*(354.3-293.8)/(0.01)

I1=5281.65 w/m^2

b)

rate of heat energy per area through radiation process is,

power P=Q/t=e*sigma*A*T^4

===>

I2=P/A=e*sigma*(T1^4-T2^4)

I2=(0.76)*(5.67*10^-8)*((354.3^4)-(293.8)^4)

I2=357.94 W/m^2

here,

through conduction process, the mug loses more amont of heat enegry,

therefore,

I1/I2=5281.65/357.94

=14.75

===> I1=(14.75)*I2 -------------> is answer

here, the mug heat loses heat by conduction is 14.75 times more quickly than by radiation.

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