Respuesta :
The amount of heat will be equal to Lm.
Where L is the latent heat of fusion and m is mass of the ice.
Latent heat of ice = 80cal/g.
So the amount of heat required here will be 35× 80cal
= 2,800 cal.
Where L is the latent heat of fusion and m is mass of the ice.
Latent heat of ice = 80cal/g.
So the amount of heat required here will be 35× 80cal
= 2,800 cal.