150 ,g of ice is mixed with 100 ,g of water at temperature 80^{circ} C . latent heat of ice is 80 ca

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10-1.Thermometry, Thermal Expansion and Calorimetry

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$150 \,g$ 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 ........... $g$

$100$

$0$

$150$

$50$

(KVPY-2012)

Solution

(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

$\Rightarrow m L=m_w s_w(\Delta T)$

$\Rightarrow m 80=100 \times 1 \times(80-0)$

$\Rightarrow m 80=100 \times 80$

$\Rightarrow m=\frac{100 \times 80}{80}$

$\Rightarrow m=100 \,g$

So, $50 \,g$ of ice does not melts.

Standard 11

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medium

$150 \,g$ 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 ........... $g$

Solution

Similar Questions

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