Question

If you know that heat lost by placing a warm piece of metal in cool water is 150 KJ, how much heat is gained by the water?

Answer 1

We know that according to conservation of energy the total energy of an isolated system is conserved.

When a warm piece of metal is placed in a container containing cool water, the system comprising the metal, cool water can only be considered isolated if we neglect the heat loss to the surrounding environment and the container which is not always the case.

But in most of the case the heat loss to the environment and container is much smaller in comparison to the heat gained by the cool water in a controlled experiment. Hence, we can safely use the conservation of energy in the system.

In general, heat lost by a system is written as negative and heat gained by a system is written as a positive value.

Hence, heat lost by our metal is

Imetal = -150 kJ

Let the heat gained by the water be $q_{water}$. Using the conservation of energy, total heat lost by the metal must equal total heat gained by the water only if we assume that heat lost to the surrounding and the calorimeter(container) is negligible.

Now, we can write

Ametal + qwater = 0 > Imetal = -(water qwater = -(-150kJ) = 150 kJ

Hence, the heat gained by the cool water must be 150 kJ.