How can the enthalpy change of a NaNO3 solution be calculated using calorimetry?

[jxs919]

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

 

[Gokul43201]

1. This question belongs in the homework & coursework subforum.

2. \Delta H = \Delta (mCT) = m_fC_fT_f - m_i C_iT_i
where f:final, i:initial

 

[jxs919]

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

 

[Gokul43201]

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).

 

[GCT]

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.