Calculate the final temperature of the solution

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Discussion Overview

The discussion revolves around a homework problem involving the dissolution of calcium chloride (CaCl2) in water and the calculation of the final temperature of the resulting solution. Participants explore the application of thermodynamic principles and specific heat capacity in this context.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant presents a calculation attempt using the heat of dissolution (ΔH) and specific heat capacity to find the final temperature of the solution.
  • Another participant questions how much CaCl2 is needed to achieve the given heat change of 81.5 kJ.
  • A different participant suggests that 1 mole of CaCl2 corresponds to the heat change, prompting further inquiry into the number of moles present in the sample.
  • Participants discuss the conversion of the mass of CaCl2 to moles and its relation to the heat value for the calculation.
  • One participant expresses gratitude for assistance after arriving at a correct answer, indicating a collaborative effort in problem-solving.

Areas of Agreement / Disagreement

The discussion shows a progression of understanding among participants, with some clarifying the relationship between mass, moles, and heat. However, there is no consensus on the final temperature calculation, as the initial attempt was deemed incorrect.

Contextual Notes

Participants do not explicitly resolve the discrepancies in the temperature calculation or the assumptions regarding heat loss and specific heat capacity. The discussion reflects ongoing exploration of the problem rather than a definitive solution.

Who May Find This Useful

Students or individuals interested in thermodynamics, calorimetry, or chemistry homework problems may find this discussion relevant.

JessicaHelena
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Homework Statement


Consider the dissolution of CaCl2.
CaCl2(s)
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Ca2+(aq) + 2 Cl-(aq) ΔH = -81.5 kJ
A 10.6-g sample of CaCl2 is dissolved in 109 g of water, with both substances at 25.0°C. Calculate the final temperature of the solution assuming no heat lost to the surroundings and assuming the solution has a specific heat capacity of 4.18 J/°C · g.

Homework Equations


Q=mcΔT

The Attempt at a Solution


I had thought Q=ΔH = -81.5kJ = -81.5 x 10^3 J
the mass would be 10.6+109 g
and since the solution overall has a heat capacity of 4.18 J/°Cg, I plugged the numbers in:
81.5 x 10^3 = (10.6+109)x4.18x(T_f-25)
Then I got T_f = 188 °C, which is wrong, but I don't understand why. I don't have much time until the assignment is due, so could someone please help me out? Thank you!
 

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How much CaCl2 do you need to get 81.5 kJ?
 
And how many moles of CaCl2 have you got?
 
@mjc123 Oh I get it now. 10.6g/M(CaCl2), multiply that by 81.5, and that's the Q value.
 
@mjc123 Thank you — I got a correct answer! Could please you help me with another problem I posted, though?
 

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