Question

Mass $m$ of a liquid $\text{A}$ is kept in a cup at a temperature of ${90}^{\circ }\text{C}$. When placed in a room having temperature of ${20}^{\circ }\text{C}$, it takes $5\text{min}$ for the temperature of the liquid to drop to ${30}^{\circ }\text{C}$. Another liquid $\text{B}$ having nearly the same density as $\text{A}$ and of mass $m$, kept in another identical cup at ${50}^{\circ }\text{C}$ takes $5\text{min}$ for its temperature to fall to ${30}^{\circ }\text{C}$ when placed in a room having temperature ${20}^{\circ }\text{C}$. If the two liquids at ${90}^{\circ }\text{C}$ and ${50}^{\circ }\text{C}$ are mixed in a calorimeter where no heat is allowed to leak, find the final temperature of the mixture. Assume that Newton's law of cooling is applicable for the given temperature ranges.

• A. ${72}^{\circ }\text{C}$
• B. ${69.7}^{\circ }\text{C}$
• C. ${66}^{\circ }\text{C}$
• D. ${52}^{\circ }\text{C}$

Answer 1

The correct option is C ${66}^{\circ }\text{C}$

Same density of two liquids means that they will occupy the same volume in the identical cups. Therefore, the surface area through which the heat leaks is same for both.
For cooling of liquid A:
$\frac{\Delta \theta }{\Delta t}=\frac{k}{s_A}({\theta }_{av}-{\theta }_0)$
$\Rightarrow \frac{(90-30)}{5}=\frac{k}{s_A}(\frac{90+30}{2}-20)$
$\Rightarrow 12=\frac{k}{s_A}\left(40\right)$
$\Rightarrow s_A=\frac{10k}{3}$
For coolling of liquid B, we have:
$\frac{50-30}{5}=\frac{k}{s_B}(\frac{50+30}{2}-20)$
$\Rightarrow 4=\frac{k}{s_B}\left(20\right)$
$\Rightarrow s_B=5k$
When the two liquids are mixed, let the final temperature of mixture be $\theta$
Heat lost by A = Heat gained by B
$\Rightarrow m.s_A(90-\theta )=m.s_B(\theta -50)$
$\Rightarrow \frac{10k}{3}(90-\theta )=5k(\theta -50)$
$\Rightarrow 180-2\theta =3\theta -150$
$\Rightarrow 5\theta =330\Rightarrow \theta ={66}^{\circ }\text{C}$