   Chapter 11, Problem 56AP

Chapter
Section
Textbook Problem

A family comes home from a long vacation with laundry to do and showers to take. The water heater has been turned off during the vacation. If the heater has a capacity of 50.0 gallons and a 4800-W heating element, how much time is required to raise the temperature of the water from 20.0°C to 60.0°C? Assume the heater is well insulated and no water is withdrawn from the tank during that time.

To determine

The time required to raise the temperature of water.

Explanation

Given info: The volume of water the heater can hold is 50.0gallons. The density of water is 1000kg/m3.

Formula to calculate the mass is,

m=ρV

• m is the mass of the water.
• ρ is the density of water.
• V is the volume of water.

Substitute 50.0 gallon for V and 1000kg/m3 fro ρ to find m.

m=(1000kg/m3)(50.0gallon×3.786liter1gallon×103m31liter)=189kg

Formula to calculate the power of the water heater is,

P=Qt

• P is the power of the water heater.
• Q is the heat supplied to the water.
• t is the time.

Formula to calculate the heat required to raise the temperature of the water is,

Q=mc(TfTi)

• m is the mass of the water
• c is the specific heat of the water.
• Tf is the final temperature.
• Ti is the initial temperature.

Substitute mc(TfTi) for Q in the formula for power and rearrange in terms of t.

P=mc(TfTi)tt=mc(TfTi)P

Thus, the expression to find the time is, t=mc(TfTi)/P

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