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Chapter :- Thermodynamics

Question :-

✨ A piece of ice with a piece of cork floating in ice-water at a temperature of 0 ° C. When the iceberg melts, the water level rises___

1. Will increase
2. It will decrease
3. There will be the same
4. Depending on the ratio of water and ice in the initial state

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Answer 1

Consider a piece of ice with a piece of cork floating in water. Let the mass of ice (frozen water in the ice) and cork be $\small\sf{m_i}$ and $\small\sf{m_c}$ respectively. Let $\small\sf{\rho_w}$ and $\small\sf{\rho_c}$ be density of water and cork respectively.

Let the volume of the combined system of ice and cork, inside the water, be $\small\sf{V_1}$ which is equal to the volume of the water displaced. As the system floats, the mass of this displaced water is equal to mass of the system, i.e.,

$\small\longrightarrow\sf{\rho_wV_1=m_i+m_c}$

$\small\longrightarrow\sf{m_i=\rho_wV_1-m_c\quad\dots(1)}$

Now the ice melted, the water in the ice gets dissolved in the water and the cork sinks in the water. The cork sinks in the water because the density of cork is greater than that of water.

$\small\longrightarrow\sf{\rho_c>\rho_w}$

Let,

$\small\longrightarrow\sf{\rho_c=\rho_w+k\quad\dots(2)}$

where k is a positive term.

Now the water rises up by the ice water and cork, so if the volume of the water risen is $\small\sf{V_2}$ then it is equal to combined volume of the ice water $\small\sf{V_i}$ and cork $\small\sf{V_c,}$ i.e.,

$\small\longrightarrow\sf{V_2=V_i+V_c}$

Since volume = mass / density,

$\small\text{ \longrightarrow\sf{V_2=\dfrac{m_i}{\rho_w}+\dfrac{m_c}{\rho_c}} }$

[Note that density of ice water is density of water.]

From (1) and (2), substituting for $\small\sf{m_i}$ and $\small\sf{\rho_c,}$

$\small\text{ \longrightarrow\sf{V_2=\dfrac{\rho_wV_1-m_c}{\rho_w}+\dfrac{m_c}{\rho_w+k}} }$

$\small\text{ \longrightarrow\sf{V_2=V_1-\dfrac{m_c}{\rho_w}+\dfrac{m_c}{\rho_w+k}} }$

$\small\text{ \longrightarrow\sf{V_2=V_1-\left(\dfrac{m_c}{\rho_w}-\dfrac{m_c}{\rho_w+k}\right)} }$

$\small\text{ \longrightarrow\sf{V_2=V_1-m_c\left(\dfrac{1}{\rho_w}-\dfrac{1}{\rho_w+k}\right)} }$

$\small\text{ \longrightarrow\sf{V_2=V_1-m_c\cdot\dfrac{\rho_w+k-\rho_w}{\rho_w(\rho_w+k)}} }$

$\small\text{ \longrightarrow\sf{V_2=V_1-\dfrac{m_ck}{\rho_w(\rho_w+k)}} }$

Since $\small\text{ \sf{\dfrac{m_ck}{\rho_w(\rho_w+k)}} }$ is a positive quantity,

$\small\Longrightarrow\sf{V_2<V_1}$

I.e., the volume of water risen after melting is less than volume of water displaced by the sinking system before melting.

Then the water level decreases during melting since the base area of the container, in which the whole system is holded up, remains unchanged.

Hence (2) is the answer.