I'm having trouble solving Delta T. The equation: The Delta H for the solution process when solid sodium hydroxide dissolves in water is 44.4kJ/mol. When a 13.9-g sample of NaOH dissolves in 250.0g of water in a coffee-cup calorimeter, the temperature increases from 23.0 C to ----C. Assume that the solution has the same specific heat as liquid water,i.e., 4.18 J/g-K.

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I'm having trouble solving Delta T. The equation: The Delta H for the solution process when solid sodium hydroxide dissolves in water is 44.4kJ/mol. When a 13.9-g sample of NaOH dissolves in 250.0g of water in a coffee-cup calorimeter, the temperature increases from 23.0 C to ----C. Assume that the solution has the same specific heat as liquid water,i.e., 4.18 J/g-K.

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I think it would be Q =mCdeltaT Solve for delta T = Q/(mC) Molar heat capacity = 75J/mol K 250g H20 x 1mol/18g = 14 mol H20 14 x 75J/mol K = 1050 J/mol K 44400 (J/mol)/ (1050 J/mol K) = 42.3 K 42.3 K = ( Temp final - temp initial) 42.2K + temp initial = temp final Have 23+273 = 296K 296K+42.3 = 338.3K or 65.3C
^You didn't take into account the moles of NaOH. In this case were using the molar enthalpy of solvation of NaOH, and the moles of NaOH. \(\sf \large q=n_{NaOH}*\Delta H_{solvation}=\dfrac{13.9~g}{39.99~g/mol}*44.4~kJ/mol\approx 1.5*10^4~J \) Rearrange the calorimetry formula: \(\sf \large q=m*C_p*(T_f-T_i)\rightarrow T_f=\dfrac{q}{m*C_p}+T_i\) Plug in values: \(\sf \large T_f=\dfrac{q}{m*C_p}+T_i=\dfrac{1.5*10^4~J}{250~g*4.18~J/^oC~g}+23^oC \approx 37.8 ^oC\)
What does the n stand for in the first line

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\(\sf \Large n\) is the symbol for moles
@aaronq I see what you did here, but why doesn't the mass of water matter in this calculation?
it does, it matters when were observing the heat from the solvation of NaOH affecting the temperature of the water. If you notice, in the calorimetry equation, the mass of the water used appears in the denominator of the rearranged formula.
@aaronq sorry! One last thing 1. okay, so you found the molar enthalpy for dissolving NaOH in H2O by multiplying the enthalpy by the number of moles of NaOH. 2. I understood how you manipulated the equation. 3. so in the process, your obviously heating the water ( I originally thought you had to use the molar heat capacity of water).. 4. I see in the equation that the number of moles cancels out! so you're left with joules.. so we don't have to (molar heat capacity of H20)..
yup, that's exactly it
@aaronq I'd give you a medal on top of a metal lol

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