An ice cube of mass 0.1 kg at $0^{o} C$ is placed in an isolated container which is at $227^{o} C$ . The specific heat S of the container varies with temperature T according to the empirical relation $S = A + B T$ , where $A = 100 c a l / k g . K$ and $B = 2 \times 10^{- 2} c a l / k g - K^{2}$ , If final temperature of container is $27^{o} C$ determine the mass of the container (Latent heat of fusion for water $= 8 \times 10^{4} c a l / k g$ , specific heat of water = $10^{3} c a l / k g - K$ )

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$4.95 k g$

$0.49 k g$

$0.295 k g$

$1.0 k g$

answer is C.

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Detailed Solution

Let m be the mass of the container.
Initial temperature of container, $T_{i} = (227 + 273) = 500 K$
and final temperature of container, $T_{f} = (27 + 273) = 300 K$
Now, heat gained by the ice cube = heat lost by the container
$∴ (0.1) (8 \times 10^{4}) + (0.1) (10^{3}) (27) = - m \int_{500}^{300} (A + B T) d T$ or $10700 = - m {[A T + \frac{B T^{2}}{2}]}_{500}^{300}$