   Chapter 11, Problem 47P

Chapter
Section
Textbook Problem

Consider two cooking pots of the same dimensions, each containing the same amount of water at the same initial temperature. The bottom of the first pot is made of copper, while the bottom of the second pot is made of aluminum. Both pots are placed on a hot surface having a temperature of 145°C. The water in the copper-bottomed pot boils away completely in 425 s. How long does it take the water in the aluminum-bottomed pot to boil away completely?

To determine
The time taken to boil away water completely from aluminum-bottomed pot.

Explanation

Given info:  The dimensions of both the pots are same, both pots are placed on hot surface of 145°C , time taken to boil away water from copper bottomed pot is 425 s.

The pots are of same dimensions. So, it holds water of equal quantity. Then the amount of energy supplied to boil away the water is also the same. Due to the different bottoms present the thermal conductivity of the pots are different.

Formula to calculate amount of heat energy transferred to the pot is,

Q=kCuA(ΔT)L(ΔtCu)

• Q is the amount of heat energy transfer to the pot,
• kCu is the thermal conductivity of copper-bottomed pot,
• ΔT is temperature difference,
• ΔtCu is the time taken to boil away water in copper bottomed flask,
•  L is the thickness of the pot,
• A is the surface area of the pot,

Formula to calculate amount of heat energy transferred to the pot is,

Q=kAlA(ΔT)L(ΔtAl)

• kAl is the thermal conductivity of copper-bottomed pot,
• ΔtAl is the time taken to boil away water in copper bottomed flask,

Equate both the energy transfer equations to obtain,

QQ=kCuA(ΔT)L(ΔtCu)kAlA(ΔT)L

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