   Chapter 17, Problem 39P

Chapter
Section
Textbook Problem

The heating element of a coffeemaker operates at 120. V and carries a current of 2.00 A. Assuming the water absorbs all the energy converted by the resistor, calculate how long it takes to heat 0.500 kg of water from room temperature (23.0°C) to the boiling point.

To determine
The time taken to heat 0.500kg of water from 23.0°C to the boiling point

Explanation

Given Info: The coffeemaker operates at 120V carries current 2.00A 0.500kg mass of water is heated from 23.0°C to the boiling point, The coffeemaker operates at 120V carries current 2.00A 0.500kg mass of water is heated from 23.0°C to the boiling point.

Explanation:

Formula to calculate the amount of heat energy is,

Q=mc(T2T1)

• Q is the amount of heat energy required to raise temperature from T1 to T2 ,
• m is the mass of water,
• c is the specific heat capacity of water,

Substitute 0.500kg for m , 4186J(kg°C)1 for c , 100°C for T2 and 23.0°C for T1 in the above equation to find Q .

Q=(0.500kg)(4186J(kg°C)1)(100°C23.0°C)=1.61×105J (I)

The amount of heat energy required to raise the temperature from 23.0°C to the boiling point is 1.61×105J

Formula to calculate the power input is,

P=IΔV

• P is the power input of heating element,
• I is the current passes,
• ΔV is the potential difference across heating element,

Substitute 2

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