A perfectly insulated stainless steel container holds 1.20 kg of 20 degree C wat

Question

A perfectly insulated stainless steel container holds 1.20 kg of 20 degree C water. Inside the container is an incandescent 100 Watt light bulb. The tungsten filament of the incandescent bulb has an emissivity of 0.25 and a surface area of 1.30 Times 10^-4 m^2. What is the temperature of the tungsten filament? Suppose that all of the energy radiated by the light bulb is absorbed by the water. How long will it take for the water to reach the boiling point? Suppose that instead of a light bulb in the container, the bottom of the container is in thermal contact with the burner of a stove that is at a temperature of 250 degree C. How long does it take for the water to reach the boiling point of water? The area of thermal contact between container and burner is 0.0315 m^2 and the thickness of the container is 0.5 cm. The thermal conductivity of stainless steel is 14 W/m-K and you can assume that a 100 degree C layer of water immediately forms at the bottom of the container. How much additional time does it take for the now boiling water to completely vaporize into steam?

Explanation / Answer

power = P = sigma*e*A*(T^4-T1^4)

100 = 5.67*10^-8*0.25*1.3*10^-4*(T2^4-293^4)

T = 2714 K

t = 2714-273 = 2441 degrees <_-------answer

+++++

b)

Q = m*C*dT = (1.2*4190*(100-20))

Q = 402240 J

time = Q/P = 4022.4 s

+++++++++++++

Q/t = K*A*dT/d

402240/t = (14*0.0315*(250-100))/0.005 = 30.4 s

d)

Q1 = mL

Q1/t1 = K*A*dT/d

1.2*2260000/t1 = (14*0.0315*(250-100))/0.005

t1 = 205 s

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