Question

If the 3.21g of NH₄NO₃ in example 5.6 were dissolved in 100.0g of water under the same conditions, how much would the temperature change? Explain your answer.

Short answer

The temperature falls by 2.3˚C if 3.21g of NH₄NO₃ are taken in 100 g of water in the same calorimeter mentioned in example 5.6.

Step-by-step solution

Heat of the reaction

In this example, through proper calculations, we find the value of the heat of the reaction to be:

q_reaction = -q_solution= -1.0 kJ

Calculation of the total mass

Since the solution is aqueous, we can proceed in terms of the water’s specific heat and mass. By substituting the mass of the water by 100 g, we now have total mass,

m = (100+3.21) g.

Calculation of the temperature

Substituting the values in the equation below.

q_solution= (c × m × ΔT)_solution

-1.0×10³J =(4.18J/g˚C)(103.2g)(∆T)

∆T = -2.3˚C

Therefore, the temperature falls by 2.3˚C. Hence, the new temperature would be 22.6˚C. Since the mass and the heat capacity of the solution are approximately equal to that of the water, the two-fold increase in the amount of water leads to a two-fold decrease in the temperature change.