Water of mass m 2 = 1 kg is contained in a copper calorimeter of mass m 1 = 1 kg . Their common temperature t = 1 0 ∘ C . Now a piece of ice of mass m 3 = 2 kg and temperature is − 1 1 ∘ C dropp

Question

Water of mass m 2 ​ = 1 kg is contained in a copper calorimeter of mass m 1 ​ = 1 kg . Their common temperature t = 1 0 ∘ C . Now a piece of ice of mass m 3 ​ = 2 kg and temperature is − 1 1 ∘ C dropped into the calorimeter. Neglecting any heat loss, the final temperature of system is. [specific heat of copper = 0.1 Kcal / kg ∘ C , specific heat of water = 1 Kcal / kg ∘ C , specific heat of ice = 0.5 Kcal / kg ∘ C , latent heat of fusion of ice = 78.7 Kcal / kg ]

TestHub · Accepted Answer

Loss in heat from calorimeter + water as temperture changes from $1 0^{\circ} C$ to $0^{\circ} C$

$= m_{1} C_{1} 10 + m_{2} C_{2} 10 = 1 \times 1 \times 10 + 1 \times 0.1 \times 10 = 11 kcal$

Gain in heat of ice as its temperature changes from $- 1 1^{\circ} C$ to $0^{\circ} C$

$= m_{3} C_{3} \times 11 = 2 \times 0.5 \times 11 = 11 kcal$

Hence ice and water will coexist at $0^{\circ} C$ without any phase change.

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