Question

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.

Answer 1

Answer:

77.08 C

Explanation:

$m$ = mass of the water = 500 g = 0.5 kg

$c$ = specific heat of water = 4186 J/(kg °C)

$\Delta T$ = Rate of change of temperature = 3 °C /min = (3/60 ) °C /s = 0.05  °C /s

$k$ = thermal conductivity of glass = 0.84

$A$ = Area of the element = 0.090 m²

$t$ = thickness of the element = 1.5 mm = 0.0015 m

$T_{i}$ = Temperature inside = 75 °C

$T_{o}$ = Temperature outside = ?

Using conservation of energy

Heat gained by water = Heat transferred through glass

$m c \Delta T = \frac{kA(T_{o} - T_{i})}{t} \\(0.5) (4186 (0.05) = \frac{(0.84)(0.090)(T_{o} - 75)}{0.0015} \\104.65 = (50.4)(T_{o} - 75)\\T_{o} = 77.08 C$