A student carried out the following reaction in a coffee cup calorimeter:

Question

A student carried out the following reaction in a coffee cup calorimeter:

 

CaCO3 +2HCl → CO2 +H2O +CaCl2

She used 4.2 g of reactant and 170.2 g of acid. The initial temperature of the system was 24.0°C and the final temperature was 26.0°C.Assuming the specific heat capacity of the solution is 4.18 J/g°C . calculate the total energy released or absorbed by the system in J, rounded to the nearest whole number. 

Answer 1

q = mC∆T

q = heat = ?

m = mass = 4.2 g + 170.2 g = 174.4 g

C = specific heat = 4.18 J/gºC

∆T = change in temperature = 26.0º - 24.0º = 2.0ºC

Solving for q:

q = (174.4 g)(4.18 J/gºC)(2.0ºC)

q = 1458 J released by the system (rounded to nearest whole number)

(note: heat is released by the system because the temperature went up from 24 to 26º so reaction is exothermic and sign of q would be negative)