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ICSE Class 10 Physics 2023 Paper

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Question : 70 of 76
Marks: +1, -0
Calculate the total amount of heat energy required to melt 200g200 \text{g} of ice at 0∘C0^{\circ} \text{C} to water at 100∘C100^{\circ} \text{C}.
(Specific latent heat of ice =336Jg−1=336 \text{Jg}^{-1}, specific heat capacity of water =4.2Jg−1 ∘C−1=4.2 \text{Jg}^{-1}\,{}^\circ\text{C}^{-1}
Solution:  
Ice at 0∘C→Q10^{\circ} \text{C} \xrightarrow[Q_1]{} water at 0∘C→Q20^{\circ} \text{C} \xrightarrow[Q_2]{} water[   at   100∘C]\text{water}[\;\text{ at }\;100^{\circ}\text{C}]
Given m=200g,  L  ice  =336Jg−1m=200 \text{g},\; L_{\;\text{ice}\;}=336 \text{Jg}^{-1}
C  water  =4.2Jg−1 ∘C−1C_{\;\text{water}\;}=4.2 \text{Jg}^{-1}\,{}^{\circ}\text{C}^{-1}
Q1=mLQ_1=m L
=200×336=200 \times 336
=67,200J=67,200 \text{J}
Q2=mcΔTQ_2=m c \Delta T
=200×4.2×(100−0)=200 \times 4.2 \times (100-0)
=84000J=84000 \text{J}
Total amount of heat=Q1+Q2\text{Total amount of heat}=Q_1+Q_2
=67200+84000=67200+84000
=151200J=151200 \text{J}
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