150 ,g of ice is mixed with 100 ,g of water a | vedclass

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Question

(d)

Let $m$ gram of ice melts and this causes final temperature of mixture at $0^{\circ} C$. Further melting of ice is then stopped. As there is no

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(d)

Let $m$ gram of ice melts and this causes final temperature of mixture at $0^{\circ} C$. Further melting of ice is then stopped. As there is no heat loss

Heat lost by ice $=$ Heat gained by water

$\Rightarrow m L=m_w s_w(\Delta T)$

$\Rightarrow m 80=100 	imes 1 	imes(80-0)$

$\Rightarrow m 80=100 	imes 80$

$\Rightarrow m=\frac{100 	imes 80}{80}$

$\Rightarrow m=100 \,g$

So, $50 \,g$ of ice does not melts.

$150 \,g$ of ice is mixed with $100 \,g$ of water at temperature $80^{\circ} C$. The latent heat of ice is $80 cal / g$ and the specific heat of water is $1 cal / g ^{\circ} C$. Assuming no heat loss to the environment, the amount of ice which does not melt is ........... $g$

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