Question

Two identical beakers labeled as $\left(X\right)$ and $\left(Y\right)$ contain $100{cm}^3$ of water each at ${20}^{o}C$. To the water in the beaker $\left(X\right)$, $100g$ of water at $0^{o}C$ was added and stirred to mix thoroughly. To the beaker $\left(Y\right)$, $100g$ of ice at $0^{o}C$ was added and stirred till it melted into water. The water in the beaker $\left(Y\right)$ will be:

• A. hotter than water in beaker $X$
• B. colder than water in beaker $X$
• C. heavier than water in beaker $X$
• D. lighter than water in beaker $X$

Answer 1

The correct option is B colder than water in beaker $X$

Here to the X, 100g of water at 0${}^{\circ }$C was added. So, the final temperature is:
$100\times$ 1 $\times$ ($T_f-0)=100\times 1\times (20-T_f$ )
$\therefore 100T_f=2000-100T_f$
$\therefore 200T_f=2000$
$\therefore T_f={10}^{\circ }C$.
In beaker Y, 100g of ice at 0${}^{\circ }$C was added. So, the final temperature is 0${}^{\circ }$C because ice melted to form water.
So, $T_f$ = 0${}^{\circ }$C
Thus, $Y$ is colder than $X$.