Question

When 50.0 g of 0.200 M NaCl at 24.1 C is added to 100.0 g M AgNO3 at 24.1 C in calorimeter, the temperature increases to 25.0 C as AgCl2 forms. Assuming the specific heat of solution and products is 4.20J/gC, calculate the approximate amount of heat in joules produced.

Answer 1

Answer:

567 J

Explanation:

Mass of NaCl = 50 grams

Mass of AgNO₃ added = 100.0 grams

Change in temperature, ΔT = (25°C - 24.1°C) = 0.9° C

Mass of the solution, m = 100 + 50 = 150 grams

Specific heat of the solution, C = 4.20 J/gC

now,

the heat released = mCΔT

on substituting the respective values,  we get

The heat released = 150 × 4.20 × 0.9 = 567 J