$$4$$ gm of steam at $$1000C$$ is added to $$20$$ gm of water at $$460C$$ in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of water in container $$540$$ $$cal/gm$$. Specific heat of water $$=$$ $$1$$ $$cal/gm-$$ $$0C$$.

Question

$$4$$ gm of steam at $$1000C$$ is added to $$20$$ gm of water at $$460C$$ in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of water in container $$540$$ $$cal/gm$$. Specific heat of water $$=$$ $$1$$ $$cal/gm-$$ $$0C$$.

Infinity Learn · Accepted Answer

Heat released by steam in conversion to water a $$1000C$$ is $$Q1$$ $$=$$ mL $$=$$ $$4$$×$$540$$ $$=$$ $$2160$$ cal. Heat required to raise  tempreature of water from $$460C$$ to $$1000C$$ is $$Q2$$ $$=$$ ms∆θ $$=$$ $$20$$×$$1$$×$$54$$ $$=$$ $$1080J$$     $$Q1$$ > $$Q2$$ and $$Q1Q2$$ $$=$$ $$2Hence$$ all steam is not converted to water only half steam shall be converted to waterFinal mass of water $$=$$ $$20$$ $$+$$ $$2$$ $$=$$ $$22$$ gm

4 gm of steam at 1000C is added to 20 gm of water at 460C in a container of negligible mass. Assuming no heat is lost to surrounding, the mass of water in container at thermal equilibrium is. Latent heat of water in container 540 cal/gm. Specific heat of water = 1 cal/gm- 0C.

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