A 400-g stainless steel 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 $0.090 \mathrm {~m} ^ { 2 }$ and is 2.0 mm thick. At a certain moment, the temperature of the water is $75 ^ { \circ } \mathrm { C }$ and it is rising at the rate of $3.0 \mathrm { C } ^ { \circ }$ per minute. What is the difference in temperature between the inside and the outside of the bottom of the tea kettle? Assume that the inner surface of the kettle is at the same temperature as the water inside. The thermal conductivity of stainless steel is $16.3 \mathrm {~W} / \mathrm { m } \cdot \mathrm { K }$ the specific heat of the steel is 448 $\mathrm { J } / \mathrm { Kg } \cdot \mathrm { K }$ and the specific heat of water is 4186 $\text { J/kg. K. }$

Question

A 400-g stainless steel 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 $0.090 \mathrm {~m} ^ { 2 }$ and is  2.0 mm thick. At a certain moment, the temperature of the water is $75 ^ { \circ } \mathrm { C }$ and it is rising at the rate of $3.0 \mathrm { C } ^ { \circ }$ per minute. What is the difference in temperature between the inside and the outside of the bottom of the tea kettle? Assume that the inner surface of the kettle is at the same temperature as the water inside. The thermal conductivity of stainless steel is $16.3 \mathrm {~W} / \mathrm { m } \cdot \mathrm { K }$ the specific heat of the steel is 448 $\mathrm { J } / \mathrm { Kg } \cdot \mathrm { K }$ and the specific heat of water is 4186 $\text { J/kg. K. }$

Quizplus · Accepted Answer

The rate of heat flow through the bottom of the kettle is:$$P = kA \frac{\Delta T}{d}$$where:* $k$ is the thermal conductivity of stainless steel* $A$ is the area of the bottom of the kettle in contact with the heating element* $\Delta T$ is the temperature difference between the inside and the outside of the bottom of the kettle* $d$ is the thickness of the bottom of the kettleThe rate of heat flow is also equal to the rate of change of the thermal energy of the water:$$P = mc_p \frac{dT}{dt}$$where:* $m$ is the mass of the water* $c_p$ is the specific heat of water* $\frac{dT}{dt}$ is the rate of change of the temperature of the waterSetting these two equations equal to each other, we get:$$kA \frac{\Delta T}{d} = mc_p \frac{dT}{dt}$$Solving for $\Delta T$, we get:$$\Delta T = \frac{mc_p}{kA} \frac{dT}{dt}$$Plugging in the given values, we get:$$\Delta T = \frac{(0.500 \text{ kg})(4186 \text{ J/kg. K})}{(16.3 \text{ W/m. K})(0.090 \text{ m}^2)} (3.0 \text{ C}^o/\text{min})$$$$\Delta T = 0.15 \text{ C}^o$$Therefore, the difference in temperature between the inside and the outside of the bottom of the tea kettle is $0.15 \mathrm { C } ^ { \circ }$.

A 400-G Stainless Steel Tea Kettle Containing 500 G 0.090 m20.090 \mathrm {~m} ^ { 2 }0.090 m2

A 400-G Stainless Steel Tea Kettle Containing 500 G $0.090 \mathrm {~m} ^ { 2 }$