Steam at 100,^oC is passed into 1.1,kg of wat | vedclass

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Question

We have,

The mass of water and the calorimeter $=(1.10+0.02)=1.12\,kg$

The specific heat capacity of the water $=4.184 	imes 10^3\,J / KgK$

Vedclass · Accepted Answer

$0.13$

Vedclass · Answer

We have,

The mass of water and the calorimeter $=(1.10+0.02)=1.12\,kg$

The specific heat capacity of the water $=4.184 	imes 10^3\,J / KgK$

Now, By using,

$Q=m c \Delta T$

Heat gained by the calorimeter and water is:

$Q =1.12 	imes 4.18 	imes 10^3 	imes 65\,J / KgK \ldots . .1$

Let the mass of the steam be $m kg$,

So, latent heat of the vaporization of water at

$100^{\circ} C =540\,cal / g =540 	imes 4.184 	imes 10^3\,J / Kg$

Therefore,

Heat lost by the steam,

$Q^{\prime}=m\left(540 	imes 4.184 	imes 10^3ight)+m\left(4.184 	imes 10^3 	imes(100-80)ight) J / K g k \quad 2$

By equating the above equations then we get,

$m (540+20)=1.12 	imes 65$

$m =\frac{1.12 	imes 65}{560}=0.13\,kg$

Steam at $100\,^oC$ is passed into $1.1\,kg$ of water contained in calorimeter of water equivalent $0.02\ kg$ at $15\,^oC$ till the temperature of the calorimeter rises to $80\,^oC$. The mass of steam condensed in kilogram is

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