If I make an ice pack with 750 mL of water, what mass of ammonium chloride must I include to get the temperature to drop from 20 degrees celsius to 5 degrees celsius. The heat of solution of the water and ammonium chloride is 14.8 kJ/mole.
To determine the mass of ammonium chloride required to achieve the desired temperature drop, we need to calculate the amount of heat energy absorbed by the mixture during the dissolving process.
The heat absorbed when dissolving a substance can be calculated using the formula:
Q = m × ΔH
Where:
Q is the heat absorbed (in joules)
m is the mass of the substance (in grams)
ΔH is the heat of solution (in J/g) or (J/mole) in this case
First, we need to convert the given volume of water in milliliters (mL) to grams (g) using the density of water (1 mL = 1 gram):
Mass of water = Volume of water × Density of water
= 750 mL × 1 g/mL
= 750 g
Next, we need to calculate the heat absorbed by the water when it changes from 20 degrees Celsius to 5 degrees Celsius. The specific heat capacity (c) of water is known to be 4.184 J/(g·°C). The temperature change (ΔT) can be calculated as:
ΔT = Final temperature - Initial temperature
= 5°C - 20°C
= -15°C
Now, we can calculate the heat absorbed by the water using the formula:
Q = m × c × ΔT
Q = 750 g × 4.184 J/(g·°C) × -15°C
Note: The negative sign indicates that the temperature is decreasing.
Now, we know that the heat absorbed by the water is equal to the heat released by the ammonium chloride. Therefore, we can equate the two equations:
Q_water = -Q_ammonium chloride
m_water × c × ΔT_water = m_ammonium chloride × ΔH
We can rearrange this equation to solve for the mass of ammonium chloride:
m_ammonium chloride = (m_water × c × ΔT_water) / ΔH
Substituting the known values:
m_ammonium chloride = (750 g × 4.184 J/(g·°C) × -15°C) / (14.8 kJ/mole × 1000 J/kJ)
Calculating the values:
m_ammonium chloride = -47.43 g / (14.8 J/mole × 1000 J/kJ)
= -0.0032 moles
Since the mass cannot be negative and moles of a substance should be positive, we disregard the negative sign and round the result to an appropriate number of significant figures:
m_ammonium chloride = 0.003 moles
Finally, to convert moles to grams, we need to use the molar mass of ammonium chloride, which is approximately 53.49 g/mol.
Mass of ammonium chloride = m_ammonium chloride × Molar mass
= 0.003 moles × 53.49 g/mol
= 0.16 g (rounded to two decimal places)
Therefore, you must include approximately 0.16 grams of ammonium chloride to achieve the desired temperature drop.