An ice cube of mass 0.1 kg at `0^@C` is placed in an isolated container which is at `227^@C`. The specific heat s of the container varies with temperature T according to the empirical relation `s=A+BT`, where `A= 100 cal//kg.K and B = 2xx 10^-2 cal//kg.K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container. (Latent heat of fusion for water = `8xx 10^4 cal//kg`, specific heat of water `=10^3 cal//kg.K`).A. `0.495 kg`B. `0.224 kg`C. `0.336 kg`D. `0.621 kg`

An ice cube of mass 0.1 kg at `0^@C` is placed in an isolated containe

1.	An ice cube of mass 0.1 kg at `0^@C` is placed in an isolated container which is at `227^@C`. The specific heat s of the container varies with temperature T according to the empirical relation `s=A+BT`, where `A= 100 cal//kg.K and B = 2xx 10^-2 cal//kg.K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container. (Latent heat of fusion for water = `8xx 10^4 cal//kg`, specific heat of water `=10^3 cal//kg.K`).A. `0.495 kg`B. `0.224 kg`C. `0.336 kg`D. `0.621 kg`
Answer» Correct Answer - A Heat recived by ice is `Q_(1)=mL+mCDelta T = 10700cal` Heat lost be the container is `Q_(2) = underset(300)overset(500)(int) m_(C) (A+BT)dT = m_(C)[AT +(BT^2)/(2)]_(300)^(500)` `=21600m_(C)` By principle of calorimetry, `Q_(1)=Q_(2)` `rArr m_(c) = 0.495 kg`.

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