Question

1 kg ice at 0°C, 1 kg water at 100°C and 1 kg steam at 100°C are mixed in an insulated closed container. After equilibrium is established, the amount of ice, water and steam are $m_i$, $m_w$ and $m_s$ kg respectively. Which one of the following options is correct? ($L_f$ = 80 cal/g, $L_v$ = 540 cal/g)

• A. $m_i = \frac{1}{3}$kg, $m_w = 1$kg, $m_s = \frac{5}{3}$kg
• B. $m_i = 0.5$ kg, $m_w = 2$ kg, $m_s = 0.5$ kg
• C. $m_i = 0$ kg, $m_w = \frac{7}{3}$kg, $m_s = \frac{2}{3}$ kg
• D. $m_i = \frac{7}{3}$kg, $m_w = 0$kg, $m_s = \frac{2}{3}$ kg

Answer 1

Answer: (C)

$m_i = 0$ kg, $m_w = \frac{7}{3}$kg, $m_s = \frac{2}{3}$ kg

Answer 2

Explanation:

1. Choose 100°C water as the energy reference (energy = 0).

   • Steam at 100°C (when condensed to water at 100°C) gives an extra 540 cal/g.
   • Ice at 0°C requires 80 cal/g to melt plus 100 cal/g to heat from 0°C to 100°C, i.e. a deficit of 180 cal/g.

2. For 1 kg (1000 g) of each:

   • Ice “energy” = –180 × 1000 = –180,000 cal.
   • Water energy = 0 cal.
   • Steam energy = +540 × 1000 = +540,000 cal.

3. Total energy = 540,000 – 180,000 = +360,000 cal.

4. Let final mass of steam = mₛ (in grams); then its energy above water = 540 mₛ (cal). For the system at 100°C, the leftover energy is carried by the steam so:
   540 mₛ = 360,000 cal → mₛ = 360,000/540 = 666.67 g = 2/3 kg.

5. Since total mass = 3 kg, the water mass m_w = 3 kg – (mass of steam) = 3 – 2/3 = 7/3 kg, and no ice remains (m_i = 0).

Answer:
m_i = 0 kg, m_w = 7/3 kg, m_s = 2/3 kg → Option (3)