A glass tea kettle containing 500 g of water is on the stove.The portion of the tea kettle that is in contact with the heating element has an area of 0.090 m² and is 1.5 mm thick.At a certain moment,the temperature of the water is 75°C,and it is rising at the rate of 3°C per minute.What is the temperature of the outside surface of the bottom of the tea kettle? Neglect the heat capacity of the kettle,and assume that the inner surface of the kettle is at the same temperature as the water inside.The thermal conductivity of glass is 0.840 W/m ∙ K and the specific heat of water is 4186 J/kg ∙ K.

Question

Quizplus · Accepted Answer

We can solve this problem using the one-dimensional heat conduction equation:
q = kA(dT/dx)
where q is the rate of heat transfer, k is the thermal conductivity, A is the cross-sectional area, dT/dx is the temperature gradient, and x is the distance from the heat source.

Since the inner surface of the kettle is at the same temperature as the water inside, we can assume that the temperature distribution is linear across the thickness of the kettle. Therefore, dT/dx = (75°C - T)/1.5 mm = (75°C - T)/0.0015 m.

The rate of heat transfer can be expressed as q = mCp(dT/dt), where m is the mass of water, Cp is the specific heat of water, and dT/dt is the rate of change of temperature of the water.

Equating q from the heat conduction equation and q from the heat transfer equation, we get:
kA(dT/dx) = mCp(dT/dt)
(0.84 W/m ∙ K)(0.090 m^2)((75°C - T)/0.0015 m) = (0.5 kg)(4186 J/kg ∙ K)(3°C/min)

Solving for T, we get:
T = 77°C

Therefore, the best choice is E) 77°C.

A Glass Tea Kettle Containing 500 G of Water Is