A well insulated Container has a mixture of ice and water kept at 0°C. The mixture has supplied heat at a constant rate of 420 watt by switching on electric heater at time t = 0 . The temperature of t

Question

A well insulated Container has a mixture of ice and water kept at 0°C. The mixture has supplied heat at a constant rate of 420 watt by switching on electric heater at time $t = 0$ . The temperature of the mixture was recorded at time $t = 150 s, 273 s$ and 378 s, and the readings were 0°C, 10°C and 20°C respectively. Find the mass of water in mixture at $t = 0$ .

Specific heat of water $= 4.2 J g^{- 1}^{\circ} C^{- 1}$ . Specific latent heat of fusion of ice $= 336 J g^{- 1}$ .

Assume that the mixture is stirred slowly to maintain a uniform temperature of its content.

TestHub · Accepted Answer

Let mass of mixture be M (in g) and mass of ice be m.

Between 273 s and 378 s (only water heated 10°C→20°C): $Ms Δ θ = P .Δ t$ ⇒ $M \times 4.2 \times 10 = 420 \times 105 ∴ M = 1050 g$

In the interval of 273 s the mass m of ice melted and mass M of water was heated from 0°C to 10°C: $mL + MS Δ θ = P Δ t$

$336 m + 1050 \times 4.2 \times 10 = 420 \times 273 \Rightarrow 8 m = 2730 - 1050 \Rightarrow m = 210 g$

mass of water = M - m = 1050 - 210 = 840 g hence (B).

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