Question

NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the temperature from 25.0 °C to 5.0 °C, how many grams of NH4NO3 should we use for every 100.0 g of water in the cold pack? Assume no heat was lost outside of cold pack, and the specific heat of the resulted solution was the same as water, or 4.184 J/(g•°C).

Answer 1

To calculate the amount of NH4NO3 needed for every 100.0 g of water in the cold pack, we can use the formula qsolution = m_solution * Cs * AT. First, convert the heat of solution from kJ/mol to J/g and then calculate the mass of NH4NO3 needed using the formula mass of NH4NO3 = (qsolution * 100.0 g) / heat of solution.

Step-by-step explanation:

To calculate the amount of NH4NO3 needed for every 100.0 g of water in the cold pack, we need to consider the heat of solution and the desired temperature decrease. The heat gained or lost by the solution can be calculated using the formula:

qsolution = m_solution * Cs * AT

Where:
qsolution = heat gained or lost by the solution
m_solution = mass of the solution (water + NH4NO3)
Cs = specific heat capacity of the solution (assumed to be the same as water, 4.184 J/(g·°C))
AT = change in temperature (final temperature - initial temperature)

First, we need to convert the given heat of solution from kJ/mol to J/g. This can be done by dividing the heat of solution by the molar mass of NH4NO3. The molar mass of NH4NO3 is obtained by adding the atomic masses of nitrogen (N), hydrogen (H), and oxygen (O):

Molar mass of NH4NO3 = (1 * atomic mass of N) + (4 * atomic mass of H) + (3 * atomic mass of O)

Once we have the heat of solution in J/g, we can calculate the amount of NH4NO3 needed using the formula:

mass of NH4NO3 = (qsolution * 100.0 g) / heat of solution

Answer 2

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Step-by-step explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y (1)

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol×
`(1mol)/(80,043g)`X

Y = 321J/g X (2)

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

X = 35,2g

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!