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Standard XII

Physics

Calorimetry

Pure water, c...

Question

Pure water, cooled to -15 $^{o}$ C, is kept in an insulated flask. Some ice is dropped into the flask. The fraction of water frozen into ice is (specific heat of ice = 0.5 cal/g- $^{o} C$ , specific heat of water = 1 cal/g- $^{o} C$ ) :

6/29

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3/35

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6/35

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2/35

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Solution

The correct option is A 6/35
Let,
M=mass of super cooled water.
m=mass of ice formed.
now,the heat evolved due to the formation of ice is used to raise the temperature of water from ${- 15}^{0} C$ to $0^{o} C$
$80 m = m \times 0.5 \times [0 - (- 15)] + (M - m) \times 1 \times 15$
$\frac{m}{M} = \frac{6}{35}$

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Pure water, cooled to -15o C, is kept in an insulated flask. Some ice is dropped into the flask. The fraction of water frozen into ice is (specific heat of ice = 0.5 cal/g-oC, specific heat of water = 1 cal/g-oC) :

The correct option is A 6/35Let,M=mass of super cooled water.m=mass of ice formed.now,the heat evolved due to the formation of ice is used to raise the temperature of water from −150C to 0oC80m=m×0.5×[0−(−15)]+(M−m)×1×15mM=635

Pure water, cooled to -15 $^{o}$ C, is kept in an insulated flask. Some ice is dropped into the flask. The fraction of water frozen into ice is (specific heat of ice = 0.5 cal/g- $^{o} C$ , specific heat of water = 1 cal/g- $^{o} C$ ) :