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Question: Using the following data, at what temperature will the wood just sink in benzene? Density of wood ...

Using the following data, at what temperature will the wood just sink in benzene?
Density of wood at 0∘C0^\circ C =8.8×102kgm3 = 8.8 \times {10^2}\dfrac{{kg}}{{{m^3}}}
Density of benzene at 0^\circ C$$$$ = 9 \times {10^2}\dfrac{{kg}}{{{m^3}}}
cubical expansivity of wood =1.4×10−4K−1 = 1.4 \times {10^{ - 4}}{K^{ - 1}}
cubical expansivity of benzene 1.2×10−3K−11.2 \times {10^{ - 3}}{K^{ - 1}}
(A)27∘C(A){27^ \circ }C
(B)21.7∘C(B){21.7^ \circ }C
(C)31∘C(C){31^ \circ }C
(C)31.7∘C(C){31.7^ \circ }C

Explanation

Solution

The condition at which an object simply sinks within a liquid is that the density of both object and liquid should be equal. So, here the wood will simply sink within benzene when the density of wood becomes equal to the density of benzene. By using the relation between density and temperature, we will find the temperature at which the density of wood becomes equal to the density of water.
Formula used:
ρ=ρ01+γθ\rho = \dfrac{{{\rho _0}}}{{1 + \gamma \theta }}, Where ρ0{\rho _0}is the initial density, γ\gamma is a coefficient of volume expansion, and θ\theta is the temperature change.

Complete step by step answer:
Given,
the density of wood, ρw=880kgm3{\rho _w} = 880\dfrac{{kg}}{{{m^3}}}
the density of benzene, ρb=900kgm3{\rho _b} = 900\dfrac{{kg}}{{{m^3}}}
coefficient of volume expansion for wood, γw=1.4×10−4/∘C{\gamma _w} = 1.4 \times {10^{ - 4}}{/^ \circ }C
coefficient of volume expansion for benzene, γb=1.2×10−3/∘C{\gamma _b} = 1.2 \times {10^{ - 3}}{/^ \circ }C
The condition at which an object simply sinks within a liquid is that the density of both object and liquid should be equal. Therefore, during this case, the wood will simply sink within benzene when the density of wood becomes equal to the density of benzene. So, we have to find the temperature at which the density of wood becomes equal to the density of water.
The relation between density and temperature is given by the subsequent equation
ρ=ρ01+γθ\rho = \dfrac{{{\rho _0}}}{{1 + \gamma \theta }}
Let us assume that the change in temperature required for obtaining equal density is θ\theta and ρ\rho be the final density of both wood and benzene.
For wood new density is given asρ=ρw1+γwθ\rho = \dfrac{{{\rho _w}}}{{1 + {\gamma _w}\theta }}
For benzene new density is given asρ=ρb1+γbθ\rho = \dfrac{{{\rho _b}}}{{1 + {\gamma _b}\theta }}
Now equate both equations.
ρb1+γbθ=ρw1+γwθ\dfrac{{{\rho _b}}}{{1 + {\gamma _b}\theta }} = \dfrac{{{\rho _w}}}{{1 + {\gamma _w}\theta }}
Now substituting the given values. We get,
900kg/m31+1.2×10−3θ=880kg/m31+1.4×10−4×θ\dfrac{{900kg/{m^3}}}{{1 + 1.2 \times {{10}^{ - 3}}\theta }} = \dfrac{{880kg/{m^3}}}{{1 + 1.4 \times {{10}^{ - 4}} \times \theta }}
900(1+1.4×10−4θ)=880(1+1.2×10−3θ)900\left( {1 + 1.4 \times {{10}^{ - 4}}\theta } \right) = 880\left( {1 + 1.2 \times {{10}^{ - 3}}\theta } \right)
900+0.126θ=880+1.056θ900 + 0.126\theta = 880 + 1.056\theta
20=0.93θ20 = 0.93\theta
θ=200.93=21.7∘C\theta = \dfrac{{20}}{{0.93}} = {21.7^ \circ }C

So, the correct answer is “Option B”.

Note: The value of temperature that we get is that the temperature change since the initial temperature was given as 0∘C{0^ \circ }C final temperature is identical as27∘C{27^ \circ }C. When any other initial temperature is mentioned ensure you add the value obtained as temperature change to the initial temperature to get the final temperature.