Calculate the amount of heat given out while 400 g of water at $30^{\circ} \mathrm{C}$ is cooled and converted into ice at $-2^{\circ} \mathrm{C}$. (i) Specific heat capacity of water $=4200 \mathrm{~J} / \mathrm{kg} \mathrm{K}$ (ii) Specific heat capacity of ice $=2100 \mathrm{~J} / \mathrm{kg} \mathrm{K}$ (iii) Specific latent heat of fusion of ice $=336000 \mathrm{~J} / \mathrm{kg}$

Explanation

To find the heat given out, we need to calculate the heat lost by water as it cools from 30°C to 0°C, the heat lost by water as it freezes into ice at 0°C, and the heat given out as ice forms at -2°C. The heat lost by water as it cools from 30°C to 0°C is given by Q1 = m × s × ΔT, where m is the mass of water, s is the specific heat capacity of water, and ΔT is the change in temperature. The heat lost by water as it freezes into ice at 0°C is given by Q2 = m × L, where L is the specific latent heat of fusion of ice. The heat given out as ice forms at -2°C is given by Q3 = m × s × ΔT, where m is the mass of ice, s is the specific heat capacity of ice, and ΔT is the change in temperature. Therefore, the total heat given out is Q = Q1 + Q2 + Q3 = m × s × ΔT + m × L + m × s × ΔT. Substituting the given values, we get Q = 400 × 4200 × (30 - 0) + 400 × 336000 + 400 × 2100 × (0 - (-2)) = 5.04 × 10^6 + 1.344 × 10^8 + 1680000 = 1.568 × 10^8 J.

Explanation

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