if u had used 40ml of water and 6g of ammonium nitrate ,rather than 20ml water n 3g ammonium nitrate. would u expect to get a larger,smaller or identical temp. change? why?
Looks like 3g/20 is the same concentration as 6g/40. So you would expect xxxxx.
To determine the expected temperature change, we need to calculate the amount of heat released during the dissolution of ammonium nitrate in water.
First, we need to find the amount of heat released per gram of ammonium nitrate dissolved. This value, known as the heat of dissolution, can be found in reference tables or online. Let's assume it is -26.4 kJ/g.
In the first scenario, where 20 mL of water and 3 g of ammonium nitrate are mixed, the total amount of heat released can be calculated as follows:
Heat released = 3 g * (-26.4 kJ/g) = -79.2 kJ
In the second scenario, where 40 mL of water and 6 g of ammonium nitrate are mixed, we can calculate the total amount of heat released in a similar way:
Heat released = 6 g * (-26.4 kJ/g) = -158.4 kJ
Comparing the two results, we see that in the second scenario, double the amount of ammonium nitrate is used, and therefore, double the amount of heat is released (-158.4 kJ vs. -79.2 kJ).
The volume of water used does not affect the amount of heat released because water has a relatively high specific heat capacity. Therefore, the change in temperature will solely depend on the amount of heat released.
Thus, using 40 mL of water and 6 g of ammonium nitrate would be expected to result in a larger temperature change compared to using 20 mL of water and 3 g of ammonium nitrate.