According to thermodynamic relationship: Q = m x s x t m = mass of the substance ( 50 ml) density of water = 1 g/ml mass of the substance ( water) = 50 ml x 1 g / ml = 50 g specific heat of water = 4.18 J/ g 0C change in temperature = ( 5- 20)0C = 150C Q = 50 g x 4.184 J/g0C x 150C = 3138 J The Heat of decomposition of NH4NO3 can be calculated as: NH4NO3 -------> NH4+ + NO3- Hrxn = { -205 + ( -132) } kJ - ( -365.6) kJ = 28.6 kJ/mol = 28600 J/mol Number of moles of NH4NO3 required = Q / Hrxn = 3136 J / 28600 J/mol = 0.1097 mol The amount of NH4NO3 required to lower the temperature of water from room temperature ( 200C to 50C ) is 0.1097 mol x 80 g/mol = 8.7776 g Solution According to thermodynamic relationship: Q = m x s x t m = mass of the substance ( 50 ml) density of water = 1 g/ml mass of the substance ( water) = 50 ml x 1 g / ml = 50 g specific heat of water = 4.18 J/ g 0C change in temperature = ( 5- 20)0C = 150C Q = 50 g x 4.184 J/g0C x 150C = 3138 J The Heat of decomposition of NH4NO3 can be calculated as: NH4NO3 -------> NH4+ + NO3- Hrxn = { -205 + ( -132) } kJ - ( -365.6) kJ = 28.6 kJ/mol = 28600 J/mol Number of moles of NH4NO3 required = Q / Hrxn = 3136 J / 28600 J/mol = 0.1097 mol The amount of NH4NO3 required to lower the temperature of water from room temperature ( 200C to 50C ) is 0.1097 mol x 80 g/mol = 8.7776 g.

1. According to thermodynamic relationship:
Q = m x s x t
m = mass of the substance ( 50 ml)
density of water = 1 g/ml
mass of the substance ( water) = 50 ml x 1 g / ml
= 50 g
specific heat of water = 4.18 J/ g 0C
change in temperature = ( 5- 20)0C
= 150C
Q = 50 g x 4.184 J/g0C x 150C
= 3138 J
The Heat of decomposition of NH4NO3 can be calculated as:

2. NH4NO3 -------> NH4+ + NO3-
Hrxn = { -205 + ( -132) } kJ - ( -365.6) kJ
= 28.6 kJ/mol
= 28600 J/mol
Number of moles of NH4NO3 required = Q / Hrxn
= 3136 J / 28600 J/mol
= 0.1097 mol
The amount of NH4NO3 required to lower the temperature of water from room temperature (
200C to 50C ) is
0.1097 mol x 80 g/mol
= 8.7776 g
Solution
According to thermodynamic relationship:
Q = m x s x t

3. m = mass of the substance ( 50 ml)
density of water = 1 g/ml
mass of the substance ( water) = 50 ml x 1 g / ml
= 50 g
specific heat of water = 4.18 J/ g 0C
change in temperature = ( 5- 20)0C
= 150C
Q = 50 g x 4.184 J/g0C x 150C
= 3138 J
The Heat of decomposition of NH4NO3 can be calculated as:
NH4NO3 -------> NH4+ + NO3-
Hrxn = { -205 + ( -132) } kJ - ( -365.6) kJ

4. = 28.6 kJ/mol
= 28600 J/mol
Number of moles of NH4NO3 required = Q / Hrxn
= 3136 J / 28600 J/mol
= 0.1097 mol
The amount of NH4NO3 required to lower the temperature of water from room temperature (
200C to 50C ) is
0.1097 mol x 80 g/mol
= 8.7776 g