InterviewSolution
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InterviewSolution
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An ice cube of mass 0.1kg 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=2xx10^-2cal//kg-K^2`. If the final temperature of the container is `27^@C`, determine the mass of the container. (Latent heat of fusion of water =`8xx10^4cal//kg`, Specific heat of water=`10^3cal//kg-K`). |
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Answer» Correct Answer - D Here the equilibrium temperature is `273+27=300K` Also according to the principle of calorimetry Heat lost by container=Heat gained by ice. Heat lost by container: Since specific heat is variable, we need to take the help of calculus to find the heat lost by the container. Let dQ be the heat lost when the temperature of the container is T. `:. dQ=mcdT` Where m is the mass of the container and `C=A+BT` is specific heat at that temperature `:. dQ=m(A+BT)dT` On integrating, we get `Q=int_500^300m(A+BT)dT=m[AT+(BT)^2/2]_500^300` `=-21600m` calorie(heat lost) Heat gained by ice This heat is to be divided into two parts (i) `0^@ iceto0^@ water` `0^@water to 27^@ water` `Q_1=mL Q_2=mcDeltaT` `=0.1xx80,000 =0.1xx10^3xx27` `=8000cal =2700cal` `:. Q_1+Q_2=8000+2700=10,700cal...(i)` Heat lost =heat gained `21600m=10,700` `rArrm=0.495kg` |
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