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ICSE Class 10 Physics 2020 Paper

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Question : 59 of 67
Marks: +1, -0
A piece of ice of mass 60 g60\ \text{g} is dropped into 140 gg of water at 50∘ C50^{\circ}\ \text{C}. Calculate the final temperature of water when all the ice has melted.
(Assume no heat is lost to the surrounding)
Specific heat capacity of water =4.2 Jg−1k−1=4.2\ \mathrm{Jg}^{-1}\mathrm{k}^{-1}
Specific latent heat of fusion of ice =336 Jg−1=336\ \mathrm{Jg}^{-1}
Solution:  
Given : mass of water (m)=140 g(m)=140\ \text{g} , initial temperature of water =50∘ C=50^{\circ}\ \text{C}
mass of ice (m′)=60 g(m')=60\ \text{g}, initial temperature of ice =0∘ C=0^{\circ}\ \text{C}.
Let the final temperature =T∘ C= T^{\circ}\ \text{C}
By the principle of method of mixture, Heat lost by water == Heat gained by ice
mcΔt  =m′L+m′CΔt′m c \Delta t\;=m' L+m' C \Delta t'
140×4.2×(50−T)  =60×336140 \times 4.2 \times (50-T)\;=60 \times 336 +60×4⋅2×(T−0)+60 \times 4 \cdot 2 \times (T-0)
588(50−T)  =20160+252T588(50-T)\;=20160+252 T
29400−588T  =20160+252T29400-588 T\;=20160+252 T
29400−20160  =588T+252T29400-20160\;=588 T+252 T
9240  =840T9240\;=840 T
  or    T=9240840=11∘C.\;\text{or}\;\; T=\frac{9240}{840}=11^{\circ}\text{C}.
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