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ICSE Class 10 Physics 2015 Paper

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Question : 67 of 74
Marks: +1, -0
A refrigerator converts 100 g100\text{ g} of water at 20∘C20^{\circ}\text{C} to ice at −10∘C-10^{\circ}\text{C} in 35 minutes.
Calculate the average rate of heat extraction in terms of watts.
Given:
Specific heat capacity of ice =2.1 J g−1 ∘C−1=2.1\text{ J}\,\text{g}^{-1}\,{}^{\circ}\text{C}^{-1}.
Specific heat capacity of water =4.2 J g−1=4.2\text{ J}\,\text{g}^{-1} ∘C−1^{\circ}\text{C}^{-1}.
Specific Latent heat of fusion of ice =336 J=336\text{ J} g−1\text{g}^{-1}
Solution:  
    Mass of water  =100 g=0.1 kg.\;\;\text{Mass of water}\;=100\text{ g}=0.1\text{ kg}.
    Temperature of water  =20∘C  .  \;\;\text{Temperature of water}\;=20^{\circ}\text{C}\;\text{.}\;
Amount of heat extracted to convert water from 20∘C20^{\circ}\text{C} to 0∘C0^{\circ}\text{C}.
Q1  =mC  water  ΔtQ_1\;=m C_{\;\text{water}\;}\Delta t
  =0.1×4200×(20−0)\;=0.1\times 4200\times (20-0)
  =8400 J\;=8400\text{ J}
Amount of heat extracted to convert water at 0∘C0^{\circ}\text{C} to 0∘C0^{\circ}\text{C} ice
Q2  =mL  ice  =0.1×336×1000Q_2\;=m L_{\;\text{ice}\;}=0.1\times 336\times 1000
  =33600 J\;=33600\text{ J}
Amount of heat extracted to convert ice at 0∘C0^{\circ}\text{C} to ice at −10∘C-10^{\circ}\text{C}.
Q3  =mC  ice  ΔtQ_3\;=m C_{\;\text{ice}\;}\Delta t
  =0.1×2.1×103×(0−(−10))\;=0.1\times 2.1\times 10^3\times (0-(-10))
  =2100 J\;=2100\text{ J}
  Total Heat  (Q)  =Q1+Q2+Q3\;\text{Total Heat}\;(Q)\;=Q_1+Q_2+Q_3
  =8400+33600+2100\;=8400+33600+2100
  =44100 J\;=44100\text{ J}
∴    Pt  =Q\therefore\;\;P t\;=Q
  or      P  =  Qt\;\text{or}\;\;\;P\;=\;\frac{Q}{t}
  Power    =  4410035×60=  441002100\;\text{Power}\;\;=\;\frac{44100}{35\times 60}=\;\frac{44100}{2100}
  =21 Watt\;=21\text{ Watt}
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