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:

• A. $1:3$
• B. $3:1$
• C. $2:1$
• D. $1:2$

Hint:

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, $\frac{s_A}{s_B} = \frac{15}{5} = 3$.

Answer 1

The Correct Option is B

Step 1: Start from the general mixing formula.
For two equal masses $m$ mixed together, the equilibrium temperature is \[ T_{mix} = \frac{s_AT_A + s_BT_B}{s_A+s_B} \] since the mass $m$ cancels out.
Step 2: Substitute the known temperatures.
With $T_A=40^\circ\text{C}$, $T_B=20^\circ\text{C}$ and $T_{mix}=35^\circ\text{C}$, \[ 35(s_A+s_B) = 40s_A + 20s_B \]
Step 3: Solve for the ratio directly. \[ 35s_A+35s_B = 40s_A+20s_B \implies 15s_B = 5s_A \implies \frac{s_A}{s_B} = 3 \]
\[ \boxed{s_A:s_B = 3:1} \]