NTSE Stage 2 SAT 2014 Question Paper

Question : 16 of 100

Marks: +1, -0

Two identical beakers labeled as (X) and (Y) contain 100

\c\m^{3}

of water each at 20°C. To the water in the beaker(X) 100 g of water at 0° C was added and stir red to mix thoroughly. To the beaker(Y) 100 g of ice at 0° C was added and stirred till it melted into water. The water in the beaker (Y) will be

hotter than water in beaker X
colder than water in beaker X
heavier than water in beaker X
lighter than water in beaker X

Validate

Solution:

In beaker (y), of 100 gm of ice at 0°C is added then final temperature = 0°C But in beaker (x), if 100 gm. of water is added then final temperature

=100×1( \T_\f -0)

=100×(20-\T_\f)=10°c

∴ Water in the beaker (y) will be colder than water in beaker (x) because of absorption of latent heat.

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