Question

There are a smallpackage of ammonium nitrate (NH4NO3) in water. When the inside bag is broken, the solute dissolves in the water. This dissolution requires energy and the water cools and freezes. I have done many experiments to find the best packs, and the market leader uses 300 g of the solute and 300 mL of water. It reaches a final temperature of -2 C. If I assume the dissolution processis so fast that it can be modelled as adiabatic, calculate the specific heat of solvation for ammonium nitrate in water.

Answer 1

we know the formula, q = ms $\Delta$ t

Assume that the initial temperature as room temperature, i.e. 25 degree C

And it is given that the adiabatic condition, so, q = 0

total heat change, m_waters_water $\Delta$ t + m_NH4NO3s_NH4NO3 $\Delta$ t = 0

300 * 4.184 * ( - 2 - 25 ) + 300 * s_NH4NO3( - 2 - 25 ) = 0

33890.4 - 8100 s_NH4NO3 = 0

s_NH4NO3 = 4.184 J / g.⁰C

SO, for the best packs like 300 g. of ammonium nitrate in 300 mL of water, the specific heat of solvation of ammonium nitrate in water is equal to the specific heat of water.