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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Thermal and Chemical Effects of Electric Current

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Thermal and Chemical Effects of Electric Current

When the room temperature is 25^ ^ ^ C , an electric kettle used...

When the room temperature is 25^{}^{{}^{\circ}} C, an electric kettle used for preparing tea takes 22 minutes to boil 44 cups of water (where 11 cup contains 200 cc200 \text{ cc} of water). Calculate the \cos t of boiling 44 cups of water, given that the \cos t of power consumption is Re 1.001.00 per unit (11 unit =1000= 1000 watt-hour).

Options

  1. A9 paise9 \text{ paise}
  2. B7 paise7 \text{ paise}
  3. C14 paise14 \text{ paise}
  4. D5 paise5 \text{ paise}

Correct answer

B. 7 paise7 \text{ paise}

Step-by-step solution

Mass of water, m=4×200 cc×1 g/cc=800 g=0.8 kgm = 4 \times 200 \text{ cc} \times 1 \text{ g/cc} = 800 \text{ g} = 0.8 \text{ kg}

Specific heat capacity of water, s = 4200 \text{ J kg}^{-1} ^{}^{{}^{\circ}}\textC^{-1}

Change in temperature, \Delta T = 100^{}^{{}^{\circ}}\text{C} - 25^{}^{{}^{\circ}}\text{C} = 75^{}^{{}^{\circ}}\text{C}

Heat required to boil the water, Q=msΔTQ = m s \Delta T

Q=0.8×4200×75=252000 JQ = 0.8 \times 4200 \times 75 = 252000 \text{ J}

Since 1 unit=1 kWh=3.6×106 J1 \text{ unit} = 1 \text{ kWh} = 3.6 \times 10^6 \text{ J}, the electrical energy consumed in units is:

Number of units = 2520003.6×106=0.07 units\dfrac{252000}{3.6 \times 10^6} = 0.07 \text{ units}

Given the \cos t of power is Re 1.001.00 (100 paise100 \text{ paise}) per unit, the total \cos t is:

\cos t = 0.07×100 paise=7 paise0.07 \times 100 \text{ paise} = 7 \text{ paise}
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