Question:
Two liquids $A$ and $B$ are at temperatures $40^\circ\text{C}$ and $20^\circ\text{C}$ respectively. When equal masses of these liquids are mixed, the temperature of the mixture is found to be $35^\circ\text{C}$. The ratio of specific heats of $A$ and $B$ is:
Two liquids $A$ and $B$ are at temperatures $40^\circ\text{C}$ and $20^\circ\text{C}$ respectively. When equal masses of these liquids are mixed, the temperature of the mixture is found to be $35^\circ\text{C}$. The ratio of specific heats of $A$ and $B$ is:
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For equal masses, the ratio of specific heats is inversely proportional to their respective temperature changes:
$\frac{s_A}{s_B} = \frac{\Delta T_B}{\Delta T_A}$.
Here, $\Delta T_B = 35 - 20 = 15^\circ\text{C}$ and $\Delta T_A = 40 - 35 = 5^\circ\text{C}$.
Thus, $\frac{s_A}{s_B} = \frac{15}{5} = 3$.
$\frac{s_A}{s_B} = \frac{\Delta T_B}{\Delta T_A}$.
Here, $\Delta T_B = 35 - 20 = 15^\circ\text{C}$ and $\Delta T_A = 40 - 35 = 5^\circ\text{C}$.
Thus, $\frac{s_A}{s_B} = \frac{15}{5} = 3$.
Updated On: Jul 22, 2026
- $1:3$
- $3:1$
- $2:1$
- $1:2$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We need to find the ratio of the specific heat capacities of two liquids, $A$ and $B$, when equal masses of these liquids at different initial temperatures are mixed to reach a given equilibrium temperature.
Step 2: Key Formula and Approach:
According to the principle of calorimetry, in an isolated system, the heat lost by the hotter substance must equal the heat gained by the colder substance:
\[ Q_{\text{lost}} = Q_{\text{gained}} \] The heat exchange equation is given by:
\[ Q = m s \Delta T \] where $m$ is the mass, $s$ is the specific heat capacity, and $\Delta T$ is the change in temperature.
Step 3: Detailed Explanation:
• Identify the parameters:
Let the mass of both liquids $A$ and $B$ be $m$.
Let the specific heat capacities of $A$ and $B$ be $s_A$ and $s_B$ respectively.
Initial temperature of liquid $A$: $T_A = 40^\circ\text{C}$
Initial temperature of liquid $B$: $T_B = 20^\circ\text{C}$
Equilibrium temperature of the mixture: $T_{\text{mix}} = 35^\circ\text{C}$
• Set up the heat exchange equation:
Liquid $A$ cools from $40^\circ\text{C}$ to $35^\circ\text{C}$, so it loses heat:
\[ Q_{\text{lost}} = m s_A \left( T_A - T_{\text{mix}} \right) = m s_A \left( 40 - 35 \right) = 5 m s_A \] Liquid $B$ warms from $20^\circ\text{C}$ to $35^\circ\text{C}$, so it gains heat:
\[ Q_{\text{gained}} = m s_B \left( T_{\text{mix}} - T_B \right) = m s_B \left( 35 - 20 \right) = 15 m s_B \]
• Equate heat lost and heat gained:
\[ 5 m s_A = 15 m s_B \] Divide both sides by $5m$:
\[ s_A = 3 s_B \] \[ \frac{s_A}{s_B} = \frac{3}{1} \]
Step 4: Final Answer:
The ratio of the specific heats of $A$ and $B$ is $3:1$, which corresponds to Option (B).
We need to find the ratio of the specific heat capacities of two liquids, $A$ and $B$, when equal masses of these liquids at different initial temperatures are mixed to reach a given equilibrium temperature.
Step 2: Key Formula and Approach:
According to the principle of calorimetry, in an isolated system, the heat lost by the hotter substance must equal the heat gained by the colder substance:
\[ Q_{\text{lost}} = Q_{\text{gained}} \] The heat exchange equation is given by:
\[ Q = m s \Delta T \] where $m$ is the mass, $s$ is the specific heat capacity, and $\Delta T$ is the change in temperature.
Step 3: Detailed Explanation:
• Identify the parameters:
Let the mass of both liquids $A$ and $B$ be $m$.
Let the specific heat capacities of $A$ and $B$ be $s_A$ and $s_B$ respectively.
Initial temperature of liquid $A$: $T_A = 40^\circ\text{C}$
Initial temperature of liquid $B$: $T_B = 20^\circ\text{C}$
Equilibrium temperature of the mixture: $T_{\text{mix}} = 35^\circ\text{C}$
• Set up the heat exchange equation:
Liquid $A$ cools from $40^\circ\text{C}$ to $35^\circ\text{C}$, so it loses heat:
\[ Q_{\text{lost}} = m s_A \left( T_A - T_{\text{mix}} \right) = m s_A \left( 40 - 35 \right) = 5 m s_A \] Liquid $B$ warms from $20^\circ\text{C}$ to $35^\circ\text{C}$, so it gains heat:
\[ Q_{\text{gained}} = m s_B \left( T_{\text{mix}} - T_B \right) = m s_B \left( 35 - 20 \right) = 15 m s_B \]
• Equate heat lost and heat gained:
\[ 5 m s_A = 15 m s_B \] Divide both sides by $5m$:
\[ s_A = 3 s_B \] \[ \frac{s_A}{s_B} = \frac{3}{1} \]
Step 4: Final Answer:
The ratio of the specific heats of $A$ and $B$ is $3:1$, which corresponds to Option (B).
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