How much heat is required to change 25.0 g of water from solid to liquid at 0 oC? Water: ΔHfus = 334 J/g; ΔHvap= 2260J/g

The heat required to change 25.0 g of water from solid ice to liquid water at 0°C  is 8350 J

Explanation:

The parameters given are

The temperature of the solid water = 0°C

The heat of fusion,   = 334 J/g

The heat of vaporization, = 2260 J/g

Mass of the solid water = 25.0 g

We note that the heat required to change a solid to a liquid is the heat of fusion, from which we have the formula for heat fusion is given as follows;

ΔH =  m ×

Therefore, we have;

ΔH =  25 g × 334 J/g = 8350 J

Which gives the heat required to change 25.0 g of water from solid ice to liquid water at 0°C  as 8350 J.

that tool would be a compass

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3