50gm ice at −10∘C is mixed with 20...
Question

# ( 50 mathrm{gm} ) ice at ( -10^{circ} mathrm{C} ) is mixed with ( 20 mathrm{gm} ) steam at ( 100^{circ} mathrm{C} ). When the mixture finally reaches its steady state inside a calorimeter of water equivalent ( 1.5 mathrm{gm} ) then : [Assume calorimeter was initially at ( 0^{circ} mathrm{C} ), Take latent heat of vaporization of water ( =540 ) cal/gm, Latent heat of fusion of water ( =80 ) cal/gm, specific heat capacity of water ( =1 ) cal/gm- ( ^{circ} mathrm{C} ), specific heat capacity of ice ( =0.5 mathrm{cal} / mathrm{gm}^{circ} mathrm{C} ) ](A) Mass of water remaining is : 67.4 gm(B) Mass of steam remaining is : ( 2.6 mathrm{gm} )(C) Mass of water remaining is : ( 67.87 mathrm{gm} )(D) Mass of steam remaining is : ( 2.13 mathrm{gm} )

JEE/Engineering Exams
Physics
Solution
64
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Ice at ( 0^{circ} mathrm{C} ) has a total latent heat of
[
mathrm{Q}=mathrm{m}^{*} mathrm{L}=540^{*} 80=43200 mathrm{cal}
]
Now for water to come at ( 0^{circ} mathrm{C} ) and it's state to
remain as water, we need to remove heat from it which is given by,
[
mathrm{Q}=mathrm{m}^{*} mathrm{Cp}^{*}(Delta mathrm{T})=540^{*} 1^{*} 80=43200 mathrm{cal}
]
So, we can see that
Heat released by water to reach from ( 80^{circ} mathrm{C} ) to ( 0^{circ} mathrm{C}= ) heat required by ice to melt into
water at ( 0^{circ} mathrm{C} )
So, the ice will melt to water and it's
temperature will be ( 0^{circ} mathrm{C} ).
Hence, the final state would be water at
( mathbf{0}^{circ} mathbf{C} )