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Ice at $$-20 ^\circ C$$ is filled upto height $$h = 10\ cm$$ in a uniform cylindrical vessel. Water at temperature $$\theta ^\circ C$$ is filled in another identical vessel up to the same height $$h= 10\ cm$$. Now, water from second vessel is poured into first vessel and it is found that level of upper surface falls through $$\Delta h = 0.5\ cm$$ when thermal equilibrium is reached. Neglecting thermal capacity of vessels, change in density of water due to change in temperature and loss of heat due to radiation, calculate initial temperature $$\theta $$ of water in $$^oC$$.
Given,

Density of water, $$\rho _{w}=1\:g\:cm^{-3}$$

Density of ice, $$\rho _{i}=0.9\:g\:cm^{-3}$$

Specific heat of water, $$s_{w} = 1\:cal/g ^\circ C$$

Specific heat of ice, $$s_{i} = 0.5\:cal/g ^\circ C$$

Specific latent heat of ice, $$L = 80\:cal/g$$


Solution

Total initial mass is $$(0.9)(10A)+(1)(10A)$$ where A is cross section area of cylinder. Now, let the height of ice in final situation be $$x$$. Then the height of water will be $$19.5-x$$.
By conservation of mass,

$$(xA)(0.9)+(A(19.5-x))(1)=(0.9)(10A)+(1)(10A)$$

Solving this, we get $$x=5cm$$
Now, heat lost by water in cooling from $$\theta$$ to $${ 0 }^{ o }C$$ is consumed in heating of ice from $${ -20 }^{ o }C$$ to $${ 0 }^{ o }C$$ and conversion of $$5cm$$ of ice into water at $${ 0 }^{ o }C$$.

Hence, $$(10A)(1)(\theta -0)=(0.9)(10A)(0.5)(0-(-20))+(5A)(0.9)(80)$$

$$(4.5)(100)=(10)(\theta )$$

Ans, $$\theta ={ 45 }^{ o }C$$

Physics

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