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Density

Water of mass...

Question

Water of mass $m_{2} = 1 k g$ is contained in a copper calorimeter of mass $m_{1} = 1 k g$ . Their common temperature $t = 10^{\circ} C$ . Now a piece of ice of mass $m_{3} = 2 k g$ and temperature at - $11^{\circ} C$ dropped into the calorimeter. Neglecting any heat loss, the final temperature of system is:
[specific heat of copper is $0.1 k c a l / k g^{\circ} C$ , specific heat of water is $1 k c a l / k g^{\circ} C$ , specific heat of ice is $0.5 k c a l / k g^{\circ} C$ , latent heat of fusion of ice is $78.7 k c a l / k g$ ]

4^{\circ} C

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- 2^{\circ} C

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0^{\circ} C

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- 4^{\circ} C

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Solution

The correct option is C $0^{\circ} C$
Formula used: $Q = m L, Q = m s Δ θ$
Given: $m_{2} = 1 k g, m_{1} = 1 k g, t = 10^{\circ} C, m_{3} = 2 k g, t_{3} = - 11^{\circ} C, s_{c} = 0.1 k c a l / g^{\circ} C,$
$s_{i c e} = 0.5 k c a l / k g^{\circ} C, L_{f} = 78.7 k c a l / k g$

Loss in heat of calorimeter + water as temperature changes from $10^{\circ} C to 0^{\circ} C$ ,
$Q_{1} = m_{1} s_{c} (10 - 0) + m_{2} s_{w} (10 - 0) = 1 \times 0.1 \times 10 + 1 \times 1 \times 10 = 11 k c a l$

Gain in heat of ice as its temperature changes from $- 11^{\circ} C to 0^{\circ} C$ ,
$Q_{2} = m_{3} s_{i c e} (0 - (- 11)) = 2 \times 0.5 \times 11 = 11 k c a l$

Comparing $Q_{1} and Q_{2}$ values,
ice and water will coexist at $0^{\circ} C$ without any phase change.

FINAL ANSWER: (a),

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