Question

5.876 g of ammonium nitrate (M.W. 80.043 g/mol) was added to a calorimeter containing 99.270 g of water at 24.5 °C. After the NH4NO3 dissolved the final temperature reached was 20.0 °C. Calculate sys for the dissolution of NH4NO3 The specific heat of water is 4.184 Jg-loc-1 and the heat capacity of the calorimeter (the "B" constant) is 15.0°C-1

Answer 1

Heat abosbed by NH4NO3 = Heat lost by water + Heat lost by calorimeter

We can use the following formula

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat (units J/g∙°C), ∆ is a symbol meaning "the change in"

∆T = change in temperature (°C Celcius)

Q = ? m = 99.27 g + 5.876 g = 105.146 g c= 4.184 J/g∙°C

∆T = 24.5 °C - 20 °C = 4.5 °C

Q =  105.146 g x 4.184 J/g∙°C x 4.5 °C =  1,979.688 Joules

Heat lost by calorimeter =   4.5 °C x 15 J/ °C = 67.5  Joules

Heat abosbed by NH4NO3 =  1,979.688 J + 67.5 J = 2047.18 Joules

Qsys for dissolution of NH4NO3 =  2047.18 Joules