Steam at 100^{\circ} {\C} is mixed with 10 {\g} of ice at 0^{\circ} {\C} so as to get water at a temperature of 40^{\circ} {\C}. The total mass of water at 40^{\circ} {\C} is (Latent heat of steam =540 {\cal} {\g}^{-1}; Latent heat of fusion of ice =80 {\cal} {\g}^{-1}; Specific heat of water =1 {\cal} {\g}^{-1}{ }^{\circ} {\C}^{-1} )

TS EAMCET Medical 10 May 2023 Shift 1 Paper

Question : 96 of 160

Marks: +1, -0

Steam at 100∘C is mixed with 10g of ice at 0∘C so as to get water at a temperature of 40∘C. The total mass of water at 40∘C is
(Latent heat of steam =540‌cal‌g−1; Latent heat of fusion of ice =80‌cal‌g−1; Specific heat of water =1‌cal‌g−1‌‌∘‌C−1 )

2 g
10 g
12 g
12.22 g

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