Gujarati
Hindi
10-1.Thermometry, Thermal Expansion and Calorimetry
hard

Pure water super cooled to $-15^o C$ is contained in a thermally insulated flask. Small amount of ice is thrown into the flask. The fraction of water frozen into ice is :

A

$3/35$

B

$6/35$

C

$6/29$

D

$2/35$

Solution

Memass of super cooled water.

$\mathrm{m}=\mathrm{mass}$ of ice formed.

now, the heat evolved due to the formation of ice is used to raise the temperature of water from $-15^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$

$80 m=m \times 0.5 \times[0-(-15)]+(M-m) \times 1 \times 15$

$\frac{m}{M}=\frac{6}{35}$

Standard 11
Physics

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