10 gm of ice cubes at 0^{circ} C are released in a tumbler (water equivalent 55 g ) at 40^{circ} C .

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10-1.Thermometry, Thermal Expansion and Calorimetry

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$10\; gm$ of ice cubes at $0\;^{\circ} C$ are released in a tumbler (water equivalent $55\; g$ ) at $40\;^{\circ} C$. Assuming that negligible heat is taken from the surroundings, the temperature(in $^o C$) of water in the tumbler becomes nearely $(L_f=80\; cal / g )$

A

$22$

B

$31$

C

$15$

D

$19$

(AIPMT-1988)

Solution

Let the final temperature be $T$

Heat required by ice $=m L+m \times s \times(T-0)$

$=10 \times 80+10 \times 1 \times T$

Heat lost by water $=55 \times(40-T)$

By using law of calorimetry, heat gained $=$ heat lost

$800+10 T=55 \times(40-T)$

$ \Rightarrow \quad T=21.54^{\circ} C =22^{\circ} C$

Standard 11

Physics

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