This is simply conservation of energy. Assume you controlled for loses of heat to, say, the container and atmosphere; using a styrofoam cup and cover is a good way to do this is a student-lab situation. Then, the total heat lost by the metal is equal to the total heat gained by the water.

The last line of calculation in your post is using the total heat lost by the metal, same as that gained by the water, and finding how much to cool/heat 1 kg. This amount applaies to the entire temperature difference through which the metal cooled. Convert to J/g, then calculate how many J/g per each degree.

This is simply conservation of energy. Assume you controlled for loses of heat to, say, the container and atmosphere; using a styrofoam cup and cover is a good way to do this is a student-lab situation. Then, the total heat lost by the metal is equal to the total heat gained by the water.

The last line of calculation in your post is using the total heat lost by the metal, same as that gained by the water, and finding how much to cool/heat 1 kg. This amount applies to the entire temperature difference through which the metal cooled. Convert to J/g, then calculate how many J/g per each degree.