300 ,gm of water at 25^{circ} C is added to 1 | vedclass

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Question

(d)

We know that latent heat of fusion of ice is $79.7 Cal$ per gram.

Let final temperature be $T$.

Then

$m _1 S \Delta T = m _2 L$

$30

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(d)

We know that latent heat of fusion of ice is $79.7 Cal$ per gram.

Let final temperature be $T$.

Then

$m _1 S \Delta T = m _2 L$

$300 	imes 1 	imes(25- T )=100 	imes 75$

$(25- T )=\frac{100 	imes 75}{300}$

$25- T =25$

$T =0^{\circ}\,C$

After that total energy left $=4.7 	imes 100$

Total mass of water $=400\,g$

Amount of water again converted into ice

$m =\frac{470}{79.7}$

$m =5.9\,g$

Thus whole mass is converted into water at $0^{\circ}\,C$, and about $5.9$ gwater is again converted into ice whose temperature is also $0^{\circ}\,C$.

After achieving the temperature of $0^{\circ}\,C$, latent heat of fusion is required firstly for conversion of water into ice then further lowering of temperature is possible. So the final temperature will be $0^{\circ}\,C$.

$300 \,gm$ of water at $25^{\circ} C$ is added to $100 \,gm$ of ice at $0^{\circ} C$. The final temperature of the mixture is ........... $^{\circ} C$

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