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. `0.595 kg`C. `0.695 kg`D. `0.795 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. `0.595 kg`C. `0.695 kg`D. `0.795 kg`
Answer» Correct Answer - A<br>Heat lost by container `= iint_(500)^(300) m_(C)(A + BT) dT`<br> `= - m_(C) [At + (BT^(2))/(2)]_(500)^(300) = 21600 m_(C)`<br> Heat gained by ice `= m_(ice) L + m_(ice)s_("water")DeltaT`<br> `= 0.1 xx 8 xx 10^(4) + 0.1 xx 10^(3) xx 27`<br> `= 10700 cal`<br> According to principal of calorimetry<br> Heat lost by container = Heat gained by ice<br> `21600 m_(C) = 10700`<br> or `m _(C) = 0.495` kg

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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. `0.595 kg`C. `0.695 kg`D. `0.795 kg`
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