An ice cube of mass $$0.1$$ kg at $$00C$$ is placed in an isolated container which is at $$2270C$$. 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 $$=$$ $$2$$×$$10-2cal/kg-K2$$. If the final temperature of the container is $$270C$$, find the mass of the container.(Latent$$ heat of fusion for water $$=$$ $$8$$×$$104$$ $$cal/kg$$, specific heat of water $$=$$ $$103$$ $$cal/kg-K)

Question

An ice cube of mass $$0.1$$ kg at $$00C$$ is placed in an isolated container which is at $$2270C$$. 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 $$=$$ $$2$$×$$10-2cal/kg-K2$$. If the final temperature of the container is $$270C$$,  find the mass of the container.(Latent$$ heat of fusion for water $$=$$ $$8$$×$$104$$ $$cal/kg$$, specific heat of water $$=$$ $$103$$ $$cal/kg-K)

Infinity Learn · Accepted Answer

Heat received by ice is    $$Q1$$ $$=$$ $$mL+mC$$∆T $$=$$ $$10700$$ cal. Heat lost be the container $$isQ2$$ $$=$$ ∫$$300500mC(A+BT)dT$$  $$=$$ mC[$$AT+BT22$$]$$300500$$                                                $$=$$ $$21600$$ mCBy principle of calorimetry, $$Q1$$ $$=$$ $$Q2$$⇒ mc $$=$$ $$0.495$$ kg

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