JEE PYQ: Thermodynamics Question 19
Question 19 - 2020 (04 Sep Shift 1)
The specific heat of water $= 4200$ J kg$^{-1}$ K$^{-1}$ and the latent heat of ice $= 3.4 \times 10^5$ J kg$^{-1}$. 100 grams of ice at 0$°$C is placed in 200 g of water at 25$°$C. The amount of ice that will melt as the temperature of the water reaches 0$°$C is close to (in grams) :
(1) 61.7
(2) 63.8
(3) 69.3
(4) 64.6
Show Answer
Answer: (1)
Solution
Here ice melts due to water. Let the amount of ice melts $= m_{\text{ice}}$
$m_w s_w \Delta\theta = m_{\text{ice}} L_{\text{ice}}$
$\therefore m_{\text{ice}} = \frac{m_w s_w \Delta\theta}{L_{\text{ice}}}$
$= \frac{0.2 \times 4200 \times 25}{3.4 \times 10^5} = 0.0617$ kg $= 61.7$ g
BETA
