Calorimetry NaNO3 Problem

In summary, the question is asking for the calculation of the change in enthalpy (delta H) for the solution process of dissolving 15.3g of NaNO3 in 100g of water in a calorimeter. The formula to use is delta H = q, where q is calculated using the mass of water, specific heat of water, and change in temperature. The specific heat of the solution can be approximated by the specific heat of water, but for accuracy, the specific heat of solutions can be found in a chemistry handbook. The mass of NaNO3 and water need to be incorporated in the calculation, and the specific heat capacity of the calorimeter must also be considered for an experimental solution.
  • #1
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lemme start off with the problem:

15.3g of NaNO3 were disssolved in 100 g of water in a calorimeter. The temperature of the water dropped from 25 C to 21.56 C. Calculate delta H for the solution process


here's what i figure:

delta H = q

q = MCT

M = mass of water

C = specific heat of water

T = change in temp


so plug and chug should get me my answer ?

i don't know why but i think i need to incorporate the mass of my NaNO3 somewhere...

help me please
 
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  • #2
1. This question belongs in the homework & coursework subforum.

2. [tex]\Delta H = \Delta (mCT) = m_fC_fT_f - m_i C_iT_i [/tex]
where f:final, i:initial
 
  • #3
uhm thanks for the formula but I'm still lost

is the massfinal.. water + substance ?

and the C final.. how do i calculate that ? i only know the specific heat of water by itself
 
  • #4
jxs919 said:
uhm thanks for the formula but I'm still lost

is the massfinal.. water + substance ?
Correct.

and the C final.. how do i calculate that ? i only know the specific heat of water by itself
I guess a reasonable approximation might be to use C(final) = C(initial) = C(water).

If you want to be more accurate, you can look up specific heats of solutions in a chemistry handbook like Lange or CRC (my guess is that you don't have to).
 
  • #5
The experimental solution to this problem would need to incorporate the specific heat capacity of the calorimeter. Yes, you need to incorporate the mass of NaNO3 and the water to find the enthalpy in terms of, let's say, kJ/"____" for the dissolution process.
 

1. What is calorimetry?

Calorimetry is the scientific measurement of heat transfer in a chemical reaction or physical process. It involves using a calorimeter, which is a device that measures the change in temperature of a substance or system.

2. What is the "NaNO3 problem" in calorimetry?

The "NaNO3 problem" refers to a common experiment in calorimetry where a known mass of sodium nitrate (NaNO3) is added to a known volume of water in a calorimeter, causing a change in temperature. The goal is to calculate the heat of dissolution of NaNO3, but the experiment can be complicated by factors such as heat loss to the surroundings and incomplete dissolution of the salt.

3. How do you calculate the heat of dissolution of NaNO3 in a calorimetry experiment?

The heat of dissolution of NaNO3 can be calculated by using the formula q = mcΔT, where q is the heat energy, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. By measuring the change in temperature and using the known values for the other variables, the heat of dissolution can be determined.

4. What are some sources of error in a calorimetry NaNO3 experiment?

Some sources of error in a calorimetry NaNO3 experiment include heat loss to the surroundings, incomplete dissolution of the salt, and incorrect measurements of mass and temperature. The use of an imperfect or poorly insulated calorimeter can also contribute to errors in the results.

5. How can the accuracy of a calorimetry NaNO3 experiment be improved?

The accuracy of a calorimetry NaNO3 experiment can be improved by using a well-insulated and calibrated calorimeter, ensuring complete dissolution of the salt, and taking multiple measurements to account for any outliers. The use of more precise measuring tools and techniques can also help improve the accuracy of the results.

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