Would the amount of heat absorbed by the dissolution in Example 5.6 appear greater, lesser, or remain the same if the heat capacity of the calorimeter were taken into account? Explain your answer

The heat required to raise the temperature of a substance is given by the formula Q = C \( \times \)m\( \times \)\(\Delta \)T, where “C” is the specific heat of the substance, “m” is the mass of the substance, and “∆T” is the change in the temperature of the substance.

The heat given off by the reaction equals that taken in by the solution.

Therefore,\({q_{reaction}} = - {q_{solution}}\).

It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction and there is no heat exchange between the calorimeter and its surroundings.

Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change:

\({q_{solution}}\) = (c × m × ΔT)\(_{solution}\)

Now, when we consider the heat capacity of the calorimeter, the equation becomes\({q_{reaction}} = - {q_{solution}} + ( - {q_{calorimeter}})\).

Since both the values have a negative sign, this implies that the heat absorbed by the dissolution would appear greater if the heat capacity of the calorimeter were taken into account. Therefore, the total heat absorbed would increase.