1. Tardigrade
  2. Question
  3. Physics
  4. The specific heat of water =4200 J kg -1 K -1 and the latent heat of ice =3.4 × 105 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 water reaches 0° C is close to (in grams) :

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 $0^{\circ} C$ is placed in $200 g$ of water at $25^{\circ} C$. The amount of ice that will melt as the temperature of water reaches $0^{\circ} C$ is close to (in grams) :

JEE MainJEE Main 2020Thermal Properties of Matter

Solution:

Here the water will provide heat for ice to melt
therefore
$m _{ w } s _{ w } \Delta \theta= m _{ ice } L _{ ice }$
$m _{ ice }=\frac{0.2 \times 4200 \times 25}{3.4 \times 10^{5}}$
$=0.0617 kg$
$=61.7 gm$
Remaining ice will remain un-melted
so correct answer is 1