Question

Latent heat of fusion of ice is 80 cal/mol. If 80 calorie heat is provided to one mole of ice then?​

Answer 1

Explanation:

3.14×108−6

7

8−6×2.64

8−15.84

−7.84

Answer 2

$∇fus(H)=80 \: cal/g \\ \\ t = 273 \: k \\ \\ kf = \frac{ {RT}^{2} }{100 \times∇fus(H) } \\ \\ kf = \frac{2 \times ( {273)}^{2} }{1,000 \times 80} \\ \\ kf = 1.863$