Question

Latent heat of fusion of ice is 80 cal/g .the heat required to melt 2 g of ice?

Answer 1

It requires 80calories to melt 1g of ice.
So,
2 g of ice melts by = 80*2 =160 calories

Answer 2

Answer : The heat required to melt 2 g of ice is 160 cal.

Solution : Given,

Latent heat of fusion = 80 cal/g

Mass of ice = 2 g

Latent heat of fusion of ice : It is defined as the amount of heat energy released or absorbed when solid ice changes to liquid at atmospheric pressure at its melting point.

Formula used :

$q=m\times \Delta H_{fusion}$

where,

q = heat required

m = mass

$\Delta H_{fusion}$ = latent heat of fusion

Now put the given values in this formula, we get

$q=2g\times 80cal/g=160cal$

Therefore, the heat required to melt 2 g of ice is 160 cal.