Thermal Properties of Matter - Result Question 45

46. $10 gm$ 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) $15^{\circ} C$

Topic 3: Newton’s Law of Cooling

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Answer:

Correct Answer: 46. (b)

Solution:

  1. (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,

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

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



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