Answer
64.8k+ views
Hint: The equation shows the relationship between the heat energy and the temperature which are different for different materials and shows that the specific heat is a value and describes how they relate to the heat energy. By using the specific heat capacity formula, the temperature is determined.
Useful formula
Specific heat capacity formula,
$Q = mc\Delta T$
Where, $Q$ is the heat energy, $m$ is the mass of the substance, $c$ is the specific heat, $\Delta T$ is the temperature difference
Complete step by step solution
the data that are given in the problem is;
temperature of the liquid $A = {12^ \circ }\,C$
temperature of the liquid $B = {19^ \circ }\,C$
temperature of the liquid $C = {28^ \circ }\,C$
All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.
1. The liquids $A$ and $B$ are mixed together. Then, the final temperature is ${16^ \circ }\,C$.
When the liquids $A$ and $B$ are mixed together,
Heat gained by the liquid $A$ is equal to the heat loss by the liquid $B$
By using specific heat capacity formula,
${Q_A} = {Q_B}$
Where, ${Q_A}$ is the heat energy of liquid $A$, ${Q_B}$ is the heat energy of liquid $B$.
${Q_A} = {Q_B}$
${m_A}{c_A}\left( {16 - 12} \right) = {m_B}{c_B}\left( {19 - 16} \right)\,.............\left( 1 \right)$
Where, ${m_A}$ is the mass of the liquid $A$, ${c_A}$ is the specific heat of the liquid $A$,${m_B}$ is the mass of the liquid $B$, ${c_B}$ is the specific heat of the liquid $B$.
All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.
Then, the equation (1) be changed as,
$m{c_A}\left( {16 - 12} \right) = m{c_B}\left( {19 - 16} \right)$
By cancelling the same term $m$ on both sides,
${c_A}\left( {16 - 12} \right) = {c_B}\left( {19 - 16} \right)$
On further simplifying,
$4{c_A} = 3{c_B}$
Therefore,
${c_B} = \dfrac{4}{3}{c_A}\,.............\left( 2 \right)$
2. The liquids $B$ and $C$ are mixed together. Then, the final temperature is ${23^ \circ }\,C$.
When the liquids $B$ and $C$ are mixed together,
Heat gained by the liquid $B$ is equal to the heat loss by the liquid $C$
By using specific heat capacity formula,
${Q_B} = {Q_C}$
Where, ${Q_B}$ is the heat energy of liquid $B$, ${Q_C}$ is the heat energy of liquid $C$.
${Q_B} = {Q_C}$
${m_B}{c_B}\left( {23 - 19} \right) = {m_C}{c_C}\left( {28 - 23} \right)\,.............\left( 3 \right)$
Where, ${m_B}$ is the mass of the liquid $B$, ${c_B}$ is the specific heat of the liquid $B$, ${m_C}$ is the mass of the liquid $C$, ${c_C}$ is the specific heat of the liquid $C$.
All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.
Then, the equation (3) be changed as,
$m{c_B}\left( {23 - 19} \right) = m{c_C}\left( {28 - 23} \right)$
By cancelling the same term $m$ on both sides,
${c_B}\left( {23 - 19} \right) = {c_C}\left( {28 - 23} \right)$
On further simplifying,
$4{c_B} = 5{c_C}$
Therefore,
${c_C} = \dfrac{4}{5}{c_B}\,................\left( 4 \right)$
Substituting the equation (3) in equation (4), then,
${c_C} = \dfrac{4}{5} \times \dfrac{4}{3}{c_A}$
On multiplying,
${c_C} = \dfrac{{16}}{{15}}{c_A}\,...............\left( 5 \right)$
3. The liquids $A$ and $C$ are mixed together. Then, the final temperature is $T$.
Heat gained by the liquid $A$ is equal to the heat loss by the liquid $C$
By using specific heat capacity formula,
${Q_A} = {Q_C}$
Where, ${Q_A}$ is the heat energy of liquid $A$, ${Q_C}$ is the heat energy of liquid $C$
${Q_A} = {Q_C}$
${m_A}{c_A}\left( {T - 12} \right) = {m_C}{c_C}\left( {28 - T} \right)\,............\left( 6 \right)$
Where,${m_A}$ is the mass of the liquid $A$, ${c_A}$ is the specific heat of the liquid $A$, ${m_C}$ is the mass of the liquid $C$, ${c_C}$ is the specific heat of the liquid $C$.
All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.
Then, the equation (6) be changed as,
$m{c_A}\left( {T - 12} \right) = m{c_C}\left( {28 - T} \right)$
By cancelling the same term $m$ on both sides,
${c_A}\left( {T - 12} \right) = {c_C}\left( {28 - T} \right)\,.................\left( 7 \right)$
On substituting the equation (5) in equation (7), then,
${c_A}\left( {T - 12} \right) = \dfrac{{16}}{{15}}{c_A} \times \left( {28 - T} \right)\,$
By cancelling the same terms on each side,
\[\left( {T - 12} \right) = \dfrac{{16}}{{15}}\left( {28 - T} \right)\,\]
On further,
\[15\left( {T - 12} \right) = 16\left( {28 - T} \right)\,\]
By multiplying,
$15T - 180 = 448 - 16T$
Takin the $T$ on one side and other terms in other side,
$15T + 16T = 448 + 180$
On further calculation,
$31T = 628$
Then,
$T = \dfrac{{628}}{{31}}$
On dividing,
$
T = {20.25^ \circ }\,C \\
T \simeq {20.3^ \circ }\,C \\
$
Thus, the temperature when $A$ and $C$ are mixed is ${20.3^ \circ }\,C$
Hence, the option (C) is correct.
Note: When the liquids are mixed together so the heat energy equation for the two liquids are equated. And all the three liquids are having the same mass then the mass value gets cancelled in the equations. The liquid $A$ and $C$ are mixed together then the final temperature is assumed as $T$.
Useful formula
Specific heat capacity formula,
$Q = mc\Delta T$
Where, $Q$ is the heat energy, $m$ is the mass of the substance, $c$ is the specific heat, $\Delta T$ is the temperature difference
Complete step by step solution
the data that are given in the problem is;
temperature of the liquid $A = {12^ \circ }\,C$
temperature of the liquid $B = {19^ \circ }\,C$
temperature of the liquid $C = {28^ \circ }\,C$
All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.
1. The liquids $A$ and $B$ are mixed together. Then, the final temperature is ${16^ \circ }\,C$.
When the liquids $A$ and $B$ are mixed together,
Heat gained by the liquid $A$ is equal to the heat loss by the liquid $B$
By using specific heat capacity formula,
${Q_A} = {Q_B}$
Where, ${Q_A}$ is the heat energy of liquid $A$, ${Q_B}$ is the heat energy of liquid $B$.
${Q_A} = {Q_B}$
${m_A}{c_A}\left( {16 - 12} \right) = {m_B}{c_B}\left( {19 - 16} \right)\,.............\left( 1 \right)$
Where, ${m_A}$ is the mass of the liquid $A$, ${c_A}$ is the specific heat of the liquid $A$,${m_B}$ is the mass of the liquid $B$, ${c_B}$ is the specific heat of the liquid $B$.
All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.
Then, the equation (1) be changed as,
$m{c_A}\left( {16 - 12} \right) = m{c_B}\left( {19 - 16} \right)$
By cancelling the same term $m$ on both sides,
${c_A}\left( {16 - 12} \right) = {c_B}\left( {19 - 16} \right)$
On further simplifying,
$4{c_A} = 3{c_B}$
Therefore,
${c_B} = \dfrac{4}{3}{c_A}\,.............\left( 2 \right)$
2. The liquids $B$ and $C$ are mixed together. Then, the final temperature is ${23^ \circ }\,C$.
When the liquids $B$ and $C$ are mixed together,
Heat gained by the liquid $B$ is equal to the heat loss by the liquid $C$
By using specific heat capacity formula,
${Q_B} = {Q_C}$
Where, ${Q_B}$ is the heat energy of liquid $B$, ${Q_C}$ is the heat energy of liquid $C$.
${Q_B} = {Q_C}$
${m_B}{c_B}\left( {23 - 19} \right) = {m_C}{c_C}\left( {28 - 23} \right)\,.............\left( 3 \right)$
Where, ${m_B}$ is the mass of the liquid $B$, ${c_B}$ is the specific heat of the liquid $B$, ${m_C}$ is the mass of the liquid $C$, ${c_C}$ is the specific heat of the liquid $C$.
All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.
Then, the equation (3) be changed as,
$m{c_B}\left( {23 - 19} \right) = m{c_C}\left( {28 - 23} \right)$
By cancelling the same term $m$ on both sides,
${c_B}\left( {23 - 19} \right) = {c_C}\left( {28 - 23} \right)$
On further simplifying,
$4{c_B} = 5{c_C}$
Therefore,
${c_C} = \dfrac{4}{5}{c_B}\,................\left( 4 \right)$
Substituting the equation (3) in equation (4), then,
${c_C} = \dfrac{4}{5} \times \dfrac{4}{3}{c_A}$
On multiplying,
${c_C} = \dfrac{{16}}{{15}}{c_A}\,...............\left( 5 \right)$
3. The liquids $A$ and $C$ are mixed together. Then, the final temperature is $T$.
Heat gained by the liquid $A$ is equal to the heat loss by the liquid $C$
By using specific heat capacity formula,
${Q_A} = {Q_C}$
Where, ${Q_A}$ is the heat energy of liquid $A$, ${Q_C}$ is the heat energy of liquid $C$
${Q_A} = {Q_C}$
${m_A}{c_A}\left( {T - 12} \right) = {m_C}{c_C}\left( {28 - T} \right)\,............\left( 6 \right)$
Where,${m_A}$ is the mass of the liquid $A$, ${c_A}$ is the specific heat of the liquid $A$, ${m_C}$ is the mass of the liquid $C$, ${c_C}$ is the specific heat of the liquid $C$.
All the three liquids are having the same mass, ${m_A} = {m_B} = {m_C} = m$.
Then, the equation (6) be changed as,
$m{c_A}\left( {T - 12} \right) = m{c_C}\left( {28 - T} \right)$
By cancelling the same term $m$ on both sides,
${c_A}\left( {T - 12} \right) = {c_C}\left( {28 - T} \right)\,.................\left( 7 \right)$
On substituting the equation (5) in equation (7), then,
${c_A}\left( {T - 12} \right) = \dfrac{{16}}{{15}}{c_A} \times \left( {28 - T} \right)\,$
By cancelling the same terms on each side,
\[\left( {T - 12} \right) = \dfrac{{16}}{{15}}\left( {28 - T} \right)\,\]
On further,
\[15\left( {T - 12} \right) = 16\left( {28 - T} \right)\,\]
By multiplying,
$15T - 180 = 448 - 16T$
Takin the $T$ on one side and other terms in other side,
$15T + 16T = 448 + 180$
On further calculation,
$31T = 628$
Then,
$T = \dfrac{{628}}{{31}}$
On dividing,
$
T = {20.25^ \circ }\,C \\
T \simeq {20.3^ \circ }\,C \\
$
Thus, the temperature when $A$ and $C$ are mixed is ${20.3^ \circ }\,C$
Hence, the option (C) is correct.
Note: When the liquids are mixed together so the heat energy equation for the two liquids are equated. And all the three liquids are having the same mass then the mass value gets cancelled in the equations. The liquid $A$ and $C$ are mixed together then the final temperature is assumed as $T$.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)