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Question

2 kg of ice at $${ -20 }^{ \circ  }C$$ is mixed with 5 kg of water at $${ 20 }^{ \circ  }C$$ in an insulating vessel having a negligible heat capacity. Calculate the final mass of water in the vessel. It is given that the specific heats of water and ice are 1 $$kcal$$ $${ kg }^{ -1 }$$ $$^{ \circ  }{ { C }^{ -1 } }$$ and 0.5 kcal $${ kg }^{ -1 }$$ $$^{ \circ  }{ { C }^{ -1 } }$$ respectively and the latent heat of fusion of ice is 80 $$kcal/kg$$.


Solution

Given: 2 kg of ice at $$−20^∘C$$ is mixed with 5 kg of water at $$20^∘C$$ in an insulating vessel having a negligible heat capacity. It is given that the specific heats of water and ice are $$1 kcal /(kg^\circ C)$$ and $$0.5 kcal/(kg^\circ C)$$ respectively and the latent heat of fusion of ice is $$80 kcal/kg$$. 
To find the final mass of water in the vessel.
Solution:
As per the given criteria,
Mass of water, $$m_w=5kg$$
Mass of the ice, $$m_i=2kg$$
Temperature of ice, $$T_i=−20^∘C$$
Temperature of water, $$T_w=20^∘C$$
Specific heat of ice, $$s_i=0.5 kcal/(kg^\circ C)$$
Specific heat of water, $$s_w=1 kcal /(kg^\circ C)$$
latent heat of fusion of ice is $$L=80 kcal/kg$$. 
Let $$M$$ be the final mass of the content
The amount of heat losed by water at 0 degreeis
$$Q_w=m_ws_s(\Delta T)=5\times 1\times20=100kcal$$
the amount of heat gained by ice to go to 0 degree is
$$Q_i=m_is_i(\Delta T)=2\times 0.5\times20=20kcal$$
heat left for absorption is
$$Q_w-Q_i=100-20=80kcal$$
This is equal to latent heat, i.e.,
$$mL=80\\\implies m=\dfrac {80}L=\dfrac {80}{80}=1kg$$
This is the mass of water from the converted ice.
So the final mass of water in the vessel is , $$M=m_w+m=5+1=6kg$$

Physics

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