# A pot containing 1.09 liter of water, initially at 18 degree C, is placed on a 1101 W electric...

## Question:

A pot containing {eq}1.09 {/eq} liter of water, initially at {eq}18 ^\circ C {/eq}, is placed on a {eq}1101\ W {/eq} electric heating element.

(a) How much heat must be supplied to the water to bring it to a boil?

(b) How much heat is necessary to boil all the water away? What is the minimum time required to boil all the water away?

## Heat Energy:

The heat energy is the energy that is required to change the temperature of the substance. The boiling temperature of the water is {eq}100^\circ {/eq}. Once it reaches the boiling temperature, then phase change takes place.

## Answer and Explanation:

**Given Data**

- Power of the heater{eq}(P) = 1101 \ W {/eq}

- Volume of the water {eq}V = 1.09 \ litre \\ V = 1.09 \times 10^{-3} \ m^{3} {/eq}

- Mass of the water {eq}m = \rho \times V \\ m = 1000 \times (1.09 \times 10^{-3}) \\ m = 1.09 \ kg {/eq}

- Initial temperature of the water {eq}T_{1} = 18^\circ {/eq}

**(a)**

Now, the heat required to boil the water would be

{eq}Q = mc \Delta T \\ Q = (1.09 \ kg)(4.18 \ kJ/kg k ) (100 -18) \\ Q = 373.61 \ kJ {/eq}

**(b)**

Now, the time required to boil the water

{eq}t = \dfrac{Q}{P} \\ t = \dfrac{373.61 \times 10^{3}\ J}{1101 \ J/s} \\ t = 339.34 \ s {/eq}

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from High School Physics: Help and Review

Chapter 17 / Lesson 12