
45gm of alcohol is needed to completely fill up a weight thermometer at ${15^0}C$. Find the weight of alcohol which will overflow when the weight thermometer is heated to ${33^0}C$.
(Given ${\gamma _a} = 121 \times {10^{ - 5}}^o{C^{ - 1}}$
(A) $0.96gm$
(B) \[0.9gm\]
(C) $1gm$
(D) $2gm$
Answer
233.1k+ views
Hint: In this question, we will use the concept of thermal apparent expansion which is the expansion of liquid in comparison to the container in which it is present and will substitute the values to obtain the answer.
Formula Used:
\[\Delta {V_{overflow}} = {V_0}{\gamma _a}\Delta T\]
Complete step by step solution:
Now, let assume for rise in temperature $\Delta T$, the final volume of the liquid to be VL which is given by
${V_L} = {V_0}(1 + {\gamma _L}\Delta T)$ where ${V_0}$ is the initial volume and ${\gamma _L}$ is the thermal expansion coefficient of liquid and $\Delta T$ is the change in temperature.
Final volume of the container be VC which will be given by
${V_C} = {V_0}(1 + {\gamma _C}\Delta T)$ where ${V_0}$is the initial volume and ${\gamma _C}$is the thermal expansion coefficient of liquid and $\Delta T$ is the change in temperature.
We know that liquids have more expansion than solids, hence ${V_{of}} = {V_L} - {V_C}$
The volume of liquid which overflow will be given by ${V_{of}} = {V_0}({\gamma _L} - {\gamma _C})\Delta T$
Here ${\gamma _L} - {\gamma _C}$ is the coefficient of apparent expansion denoted by ${\gamma _a}$, which is given in the question,
So, the expression of volume of overflown liquid is ${V_{of}} = {V_0}({\gamma _a})\Delta T$
We know that volume of any substance is the ratio of mass per density. Let us assume that mass of overflown liquid be mof and the density be $\rho $
Now let us put the given values in the above equation,
$\dfrac{{{m_{of}}}}{\rho } = \dfrac{{{m_0}}}{\rho }({\gamma _a})\Delta T \Rightarrow {m_{of}} = {m_0}({\gamma _a})\Delta T$, the initial mass is given to be 45gm
${m_{of}} = 45 \times 121 \times {10^{ - 5}} \times (33 - 15) = 0.98gm \approx 1gm$
So, the correct option is C.
Note:
The volume of liquid overflow can also be called the expansion of liquid with respect to the container as the liquid that is coming out is expanded extra. The sum of apparent expansion by liquid and the volume expansion by container is the real expansion of liquid. Apparent expansion is always less than the actual expansion.
Formula Used:
\[\Delta {V_{overflow}} = {V_0}{\gamma _a}\Delta T\]
Complete step by step solution:
Now, let assume for rise in temperature $\Delta T$, the final volume of the liquid to be VL which is given by
${V_L} = {V_0}(1 + {\gamma _L}\Delta T)$ where ${V_0}$ is the initial volume and ${\gamma _L}$ is the thermal expansion coefficient of liquid and $\Delta T$ is the change in temperature.
Final volume of the container be VC which will be given by
${V_C} = {V_0}(1 + {\gamma _C}\Delta T)$ where ${V_0}$is the initial volume and ${\gamma _C}$is the thermal expansion coefficient of liquid and $\Delta T$ is the change in temperature.
We know that liquids have more expansion than solids, hence ${V_{of}} = {V_L} - {V_C}$
The volume of liquid which overflow will be given by ${V_{of}} = {V_0}({\gamma _L} - {\gamma _C})\Delta T$
Here ${\gamma _L} - {\gamma _C}$ is the coefficient of apparent expansion denoted by ${\gamma _a}$, which is given in the question,
So, the expression of volume of overflown liquid is ${V_{of}} = {V_0}({\gamma _a})\Delta T$
We know that volume of any substance is the ratio of mass per density. Let us assume that mass of overflown liquid be mof and the density be $\rho $
Now let us put the given values in the above equation,
$\dfrac{{{m_{of}}}}{\rho } = \dfrac{{{m_0}}}{\rho }({\gamma _a})\Delta T \Rightarrow {m_{of}} = {m_0}({\gamma _a})\Delta T$, the initial mass is given to be 45gm
${m_{of}} = 45 \times 121 \times {10^{ - 5}} \times (33 - 15) = 0.98gm \approx 1gm$
So, the correct option is C.
Note:
The volume of liquid overflow can also be called the expansion of liquid with respect to the container as the liquid that is coming out is expanded extra. The sum of apparent expansion by liquid and the volume expansion by container is the real expansion of liquid. Apparent expansion is always less than the actual expansion.
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