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ICSE Class 10 Physics 2014 Paper

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Question : 55 of 66
Marks: +1, -0
Heat energy is supplied at a constant rate to 100 g100\ \mathrm{g} of ice at 0∘ C0^{\circ}\ \mathrm{C}. The ice is converted into water at 0∘ C0^{\circ}\ \mathrm{C} in 2 minutes. How much time will be required to raise the temperature of water from 0∘ C0^{\circ}\ \mathrm{C} to 20∘ C20^{\circ}\ \mathrm{C} ? [Given : sp. heat capacity of water =4.2 J g−1 ∘C−1=4.2\ \mathrm{J}\ \mathrm{g}^{-1}\,{}^{\circ}\mathrm{C}^{-1}, sp. latent heat of ice =336 J g−1]=336\ \mathrm{J}\ \mathrm{g}^{-1}].
Solution:  
Given : m=100 g, t=2m=100\ \mathrm{g},\ t=2 minutes =2×60=2 \times 60 sec.
Heat energy taken by ice at 0∘ C0^{\circ}\ \mathrm{C} to convert to water at 0∘ C0^{\circ}\ \mathrm{C}.
  Q=mL=100×336=33600 J\; Q=m L=100 \times 336=33600\ \mathrm{J}
  ∴    P=  Qt=  336002×60=280 J/s\; \therefore \;\; P=\; \frac{Q}{t}=\; \frac{33600}{2 \times 60}=280\ \mathrm{J}/\mathrm{s}
The heat energy required to convert water at 0∘ C0^{\circ}\ \mathrm{C} to 20∘ C20^{\circ}\ \mathrm{C} .
Q  =mcΔt=100×4.2×20=8400 JQ\;=m c \Delta t=100 \times 4.2 \times 20=8400\ \mathrm{J}
Q  =P×tQ\;= P \times t
8400  =280×t8400\;=280 \times t
t  =30 sec=0.5 min.t\;=30\ \mathrm{sec}=0.5\ \mathrm{min} .
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