Question

Equal masses of two liquids $\text{A}$ and $\text{B}$ contained in vessels of negligible heat capacities are supplied heat at the same rate. The temperature time graphs for the two liquids are shown in the figure.
If $S$ represents specific heat and $L$ represents latent heat of liquid, then -

• A. $S_A>S_B;L_A<L_B$
• B. $S_A>S_B;L_A>L_B$
• C. $S_A<S_B;L_A<L_B$
• D. $S_A<S_B;L_A>L_B$

Answer 1

The correct option is D $S_A<S_B;L_A>L_B$

From the given $T-t$ graph, we observe that the temperature of $\text{A}$ rises faster than the temperature of $\text{B}.$

Therefore, specific heat of $\text{A}$ is less than that of $\text{B}.$

$\because S=\frac{Q}{m\Delta T}[\because m\text{and}Q=\text{same}]$

$\Rightarrow S_A<S_B$

The horizontal portion of the graph represents the phase change, conversion of liquid into vapours.

The length of horizontal portion is greater for liquid $\text{A}$ compared to $\text{B}$

$\because Q=m{L}_f[\because m=\text{same}]$

$\Rightarrow L_A>L_B$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.

Why this question? Tip: Always remember that the temperature of system keeps increasing by heat supply, if there is no phase change is occurring.