An immersion heater used to boil water for a single cup of tea plugs into a 120

Question

An immersion heater used to boil water for a single cup of tea plugs into a 120 V outlet and is rated at500 W . Suppose your super-size, super-insulated tea mug contains 400 g of water at a temperature of 18 C. You can ignore the energy needed to raise the temperature of the mug and the heater itself.

Part A

What is the resistance of the heater?

Express your answer to two significant figures and include the appropriate units.

Part B

How long will this heater take to bring the water to a boil?

Express your answer using two significant figures.

Explanation / Answer

given applied voltage = V = 120 v

power = 500 W

but power = v^2/R

R = resistance

R = v^2/P = 120^2/500 = 28.8 ohms

----------

part(B)

heat required to raise the temperature of water from 18 to 100

Q = m*sw*dT

sw = specfic heat of water = 4186 J/kg

m = mass of thewater = 0.4 kg

Q = 0.4*4186*(100-18) = 137300.8 J

time taken= t = Q/P

t = 274.61 s   or 4.58 minutes

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.