Question

calculate the latent heat of fusion of ice experimentally.(The specific heat of water is 4200 Jkg-1K-1​

Answer 1

Answer:

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) (1) 69.3 (2) 63.8 (3) 64.6 (4) 61.7Read more on Sarthaks.com - https://www.sarthaks.com/903694/the-specific-heat-of-water-4200-kg-and-the-latent-heat-of-ice-4-10-5-kg-100-grams-of-ice-at-0c-is