JEE PYQ: Thermodynamics Question 35
Question 35 - 2019 (12 Apr Shift 1)
When $M_1$ gram of ice at $-10$ $°$C (Specific heat $= 0.5$ cal g$^{-1}$ $°$C$^{-1}$) is added to $M_2$ gram of water at 50$°$C, finally no ice is left and the water is at 0$°$C. The value of latent heat of ice, in cal g$^{-1}$ is:
(1) $\frac{50M_2}{M_1} - 5$
(2) $\frac{5M_1}{M_2} - 50$
(3) $\frac{50M_2}{M_1}$
(4) $\frac{5M_2}{M_1} - 5$
Show Answer
Answer: (1)
Solution
$M_1 C_{\text{ice}} \times (10) + M_1 L = M_2 C_w (50)$
$M_1 \times C_{\text{ice}}(-0.5) \times 10 + M_1 L = M_2 \times 1 \times 50$
$\Rightarrow L = \frac{50M_2}{M_1} - 5$
BETA
