If the heat of fusion of water is 3.4 × 102 j/g, the amount of heat energy required to change 15.0 grams of ice at 0°c to 15.0 grams of water at 0°c is

If the heat of fusion of water is 3.4 × 10^2 joule/gram, the heat energy amount required to change 15.0 grams of ice at 0°c to 15.0 grams of water at 0°c is 5100 joule (5100 = 3.4 x 10^2 x 15).

The problem above can be solved by using the heat of fusion formula. The formula for calculating the heat of fusion described as C = q/m, where: C is the heat of fusion, q is the amount of heat energy required, and m is the mass of the object.

We first assume that this gas is an ideal gas where it follows the ideal gas equation. the said equation is expressed as: pv = nrt. from this equation, we can predict the changes in the pressure, volume and temperature. if the volume and the temperature of this gas is doubled, then the pressure still stays the same.