Heat Absorption

Question

A generic solid, X, has a molar mass of 77.3 g/mol In a constant‑pressure calorimeter, 41.4 g of X is dissolved in 227 g of water at 23.00 °C.

X(s)⟶X(aq)X(s)⟶X(aq)

The temperature of the resulting solution rises to 26.60 ∘C. Assume the solution has the same specific heat as water, 4.184 J/(g·°C), and that there is negligible heat loss to the surroundings.

How much heat was absorbed by the solution?

What is the enthalpy of the reaction?

Answer 1

We can use the equation q = mC∆T

q = heat = ?

m = mass = 41.4 g + 227 g = 268.4 g

C = specific heat = 4.184 J/gº

∆T = change in temperature = 26.60 - 23.00 = 3.60º

Solving for q, we have...

q = (268.4 g)(4.184 J/gº)(3.6) = 4042 J (4.04kJ) of heat absorbed by the solution

Enthalpy of reaction: = ∆H_rxn = -4.04 kJ

Molar enthalpy: 41.4 g x 1 mol/77.3 g = 0.536 moles: -4.04kJ/0.536 mol = 7.54 kJ/mol