SATHEE: Thermal Properties of Matter

Thermal Properties of Matter - Result Question 45

46. $10 g m$ 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 $(L = 80 c a l / g)$ [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:

(b) Let the final temperature be $T$

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

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

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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Thermal Properties of Matter - Result Question 45

46. 10gm of ice cubes at 0∘C are released in a tumbler (water equivalent 55g ) at 40∘C.

Topic 3: Newton’s Law of Cooling

Answer:

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46. $10 g m$ of ice cubes at $0^{\circ} C$ are released in a tumbler (water equivalent $55 g$ ) at $40^{\circ} C$ .