Thermal Properties Of Matter Ques 41
41. $10$ $ 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 of water in the tumbler becomes nearly $(L=80 $ $cal / g)$
[1988]
(a) $31^{\circ} C$
(b) $22^{\circ} C$
(c) $19^{\circ} C$
(d) $15^{\circ} C$
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Answer:
Correct Answer: 41.(b)
Solution:
- (b) Let the final temperature be $T$
Heat gained by ice $=mL+m \times s \times(T-0)$
$ =10 \times 80+10 \times 1 \times T $
Heat lost by water $=55 \times 1 \times(40-T)$
By using law of calorimetery,
$800+10 T=55 \times(40-T)$
$\Rightarrow T=21.54^{\circ} C=22^{\circ} C$
BETA
