Question

Ice at -20°C is filled upto height h = 10 cm in a uniform cylindrical vessel. Water at temperature 0°C is filled in another identical vessel upto the same height h= 10 cm. Now, water from second vessel is poured into first vessel and it is found that level of upper surface falls through ∆h=0.5 cm when thermal equilibrium is reached. Neglecting thermal capacity of vessels, change in density of water due to change in temperature and loss of heat due to radiation, calculate initial temperature of water. Given,

Density of water: $\rho_w$ = 1 gm cm$^{-3}$ Specific heat of water: $s_w$ = 1 cal/gm °C Specific latent heat of ice : L = 80 cal/gm

Density of ice : $\rho_i$ = 0.9 gm/cm³ Specific heat of ice: $s_i$ = 0.5 cal/gm°C

• A. 12.6°C
• B. 20°C
• C. 45°C
• D. 765°C

Answer 1

Answer: (C)

45°C

Answer 2

The problem statement contains inconsistencies, but a plausible interpretation can be made to derive an answer.

1. Initial Quantities: Let $A$ be the cross-sectional area of the cylindrical vessel. Initial height $h = 10$ cm. Mass of ice $m_i = \rho_i \times A \times h = 0.9 \text{ gm/cm}^3 \times A \times 10 \text{ cm} = 9A \text{ gm}$. Mass of water poured $m_w = \rho_w \times A \times h = 1 \text{ gm/cm}^3 \times A \times 10 \text{ cm} = 10A \text{ gm}$.

2. Volume Change Interpretation: The initial volume occupied by ice is $V_i = A \times h = 10A \text{ cm}^3$. The level of the upper surface falls by $\Delta h = 0.5$ cm, so the final height is $h_f = h - \Delta h = 10 - 0.5 = 9.5$ cm. The final volume of the mixture is $V_f = A \times h_f = 9.5A \text{ cm}^3$. The reduction in volume is $\Delta V = V_i - V_f = 10A - 9.5A = 0.5A \text{ cm}^3$. This volume reduction is due to the melting of ice. The volume change per gram of ice melted is given by the difference in specific volumes: $\Delta V_{per\_gram} = \frac{1}{\rho_w} - \frac{1}{\rho_i} = \frac{1 \text{ cm}^3}{\text{gm}} - \frac{1}{0.9 \text{ gm/cm}^3} = 1 - \frac{10}{9} = -\frac{1}{9} \text{ cm}^3/\text{gm}$. The negative sign indicates a volume decrease upon melting. So, the magnitude of volume reduction per gram is $1/9 \text{ cm}^3/\text{gm}$. Let $m_{melt}$ be the mass of ice that melts. Total volume reduction = $m_{melt} \times \frac{1}{9}$. Equating this to the observed volume reduction: $m_{melt} \times \frac{1}{9} = 0.5A$ $m_{melt} = 0.5A \times 9 = 4.5A \text{ gm}$.

3. Calorimetry: We assume the final thermal equilibrium temperature is $T_f = 0$°C, as this is the only temperature where ice and water coexist in equilibrium, and the volume change suggests partial melting. The initial temperature of the water is $T_w$.

   Heat gained by ice to reach 0°C: $Q_{ice\_heat} = m_i \times s_i \times (0 - (-20)) = 9A \text{ gm} \times 0.5 \text{ cal/gm°C} \times 20 \text{ °C} = 90A \text{ cal}$.

   Heat gained by melting ice: $Q_{melt} = m_{melt} \times L = 4.5A \text{ gm} \times 80 \text{ cal/gm} = 360A \text{ cal}$.

   Total heat gained by the ice = $Q_{ice\_heat} + Q_{melt} = 90A + 360A = 450A \text{ cal}$.

   Heat lost by water to reach 0°C: $Q_{water\_lost} = m_w \times s_w \times (T_w - 0) = 10A \text{ gm} \times 1 \text{ cal/gm°C} \times T_w \text{ °C} = 10A T_w \text{ cal}$.

   By the principle of calorimetry, heat lost equals heat gained: $Q_{water\_lost} = Q_{ice\_heat} + Q_{melt}$ $10A T_w = 450A$ $T_w = \frac{450A}{10A} = 45$ °C.

   Note: This solution assumes the final temperature is 0°C and interprets the volume change directly to find the mass of melted ice. Checking the final volume with this $m_{melt}$ leads to inconsistencies, indicating the problem statement itself is flawed. However, this interpretation yields a plausible numerical answer.