
The temperature of equal masses of three different liquids $A$, $B$ and $C$ are ${12^ \circ }\,C$, ${19^ \circ }\,C$ and ${28^ \circ }\,C$ respectively. The temperature when $A$ and $B$ are mixed is ${16^ \circ }\,C$ and when $B$ and $C$ are mixed is ${23^ \circ }\,C$. What is the temperature when $A$ and $C$ are mixed?
(A) ${18.2^ \circ }\,C$
(B) ${22^ \circ }\,C$
(C) ${20.3^ \circ }\,C$
(D) ${24.2^ \circ }\,C$
Answer
123.3k+ 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
Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

Class 11 JEE Main Physics Mock Test 2025

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
