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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Calorimetry

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Calorimetry

On a winter day, the tap water temperature is 20^ ^ ^ C while the...

On a winter day, the tap water temperature is 20^{}^{{}^{\circ}}\text{C} while the room temperature is 5^{}^{{}^{\circ}}\text{C}. Water is stored in a tank of capacity 0.5 m30.5 \text{ m}^3 for household use. If it were possible to utilize the heat liberated by the water to vertically lift a 10 kg10 \text{ kg} mass, how high can it be lifted as the water reaches room temperature? Take g=10 m s−2g = 10 \text{ m s}^{-2}.

Options

  1. A3150 km3150 \text{ km}
  2. B315 km315 \text{ km}
  3. C3.15 km3.15 \text{ km}
  4. D31.5 km31.5 \text{ km}

Correct answer

B. 315 km315 \text{ km}

Step-by-step solution

Volume of water, V=0.5 m3V = 0.5 \text{ m}^3

Mass of water, M=Vρ=0.5×1000=500 kgM = V \rho = 0.5 \times 1000 = 500 \text{ kg}

Heat liberated by water as it cools from 20^{}^{{}^{\circ}}\text{C} to 5^{}^{{}^{\circ}}\text{C} is Q=MsΔTQ = M s \Delta T

Substituting the standard specific heat of water s = 4200 \text{ J kg}^{-1} ^{}^{{}^{\circ}}\textC^{-1}, we get:

Q=500×4200×(20−5)=3.15×107 JQ = 500 \times 4200 \times (20 - 5) = 3.15 \times 10^7 \text{ J}

This heat is utilized to lift a mass m=10 kgm = 10 \text{ kg} to a height hh. The work done is equal to the increase in potential energy.

mgh=Qmgh = Q

10×10×h=3.15×10710 \times 10 \times h = 3.15 \times 10^7

100h=3.15×107100 h = 3.15 \times 10^7

h=3.15×105 m=315 kmh = 3.15 \times 10^5 \text{ m} = 315 \text{ km}
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