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ICSE Class 10 Physics 2016 Paper

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Question : 6 of 73
Marks: +1, -0
Calculate the mass of ice required to lower the temperature of 300 g300\,\mathrm{g} of water at 40∘ C40^{\circ}\,\mathrm{C} to water at 0∘ C0^{\circ}\,\mathrm{C}.
(Specific latent heat of ice =336 J/g=336\,\mathrm{J}/\mathrm{g},
Specific heat capacity of water =4.2 J/g∘C=4.2\,\mathrm{J}/{\mathrm{g}^{\circ}\mathrm{C}} ).
Solution:  
Heat extracted by ice at 0∘ C0^{\circ}\,\mathrm{C} to melt == Heat given by water to reach 0∘ C0^{\circ}\,\mathrm{C} from 40∘ C40^{\circ}\,\mathrm{C}.
ML=mcθML = m c \theta
Where, θ=\theta = change in temperature of water
L=L = Specific Latent heat of ice
c=c = Specific heat capacity of water
M=M = Mass of ice required
m=m = mass of water
Heat gained by ice == Heat lost by water
M×336  =300×4.2×(40−0)M \times 336 \; = 300 \times 4.2 \times (40-0)
M  =  300×4.2×40336=150 gmM \; = \; \frac{300 \times 4.2 \times 40}{336} = 150 \, \mathrm{gm}
∴\therefore Ice is required to lower the temperature of water =150 gm=150\,\mathrm{gm}.
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