CaCl2(s) -> Ca(aq) + 2Cl(aq) ΔH_rxn = +198.0 KJ
"If that g sample of calcium chloride was dissolved in 1000.0g of water, with both substances originally at 25.0C, what would be the final temperature of the solution be? Assume no loss of heat to the surroundings and that the entire solution has a specific heat capacity of 4.184J/gC."
I'm not quite sure if there are three q's (one for CaCl2, one for H20 and one for the heat absorbed), or just 2 q's which is juse about CaCl2 and H20.). Also getting an answer between 0 to 25C is impossible.
For additional information, the amount of heat that would be absorbed for the same amount of CaCl2 is +40.2KJ.
Ok, if the dissolving gave off 176kj, that will heat the solution
176kj=1000*4.18*deltaTemp
deltTemp=44 C about
final temp, add intial temp 25
Oh god, I messed up my question. Apologizes. It was 22.33g calcium chloride dissolved in 1000.0g of water.
So, from what you're saying:
(Heat Absorbed is 40.2KJ, need to convert that to J or g/J?) = 1000g * 4.184 (T_f - 25)
1) 40200J = (1000g + 22.33g) * 4.184J/gC * (T_f - 25)
2) 40200J = (1022.33) * 4.184J/gC * (T_f - 25)
3) 40200J = 4277.42872 * (T_f - 25)
3) 40200J = 4277.42872T_f - 106935.718
4) 40200J + 106935.718 = 4277.42872T_f - 106935.718 + 106935.718
5) 147135.718 = 4277.42872T_f
6) 147135.718/4277.42872 = 4277.42872T_f/4277.42872
7)34.3981694685 = T_f
=> T_f ~= 34.40C
If I am wrong, please tell me.
Thank you bobpursley, I was unable to get that question right on my homework. ;)
To calculate the final temperature of the solution, you need to consider the heat gained by the water and the heat released by the dissolution of calcium chloride.
First, let's calculate the heat gained by the water. The heat gained or lost by an object can be calculated using the equation:
q = m * c * ΔT
where:
q is the heat gained or lost (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity (in J/g°C), and
ΔT is the change in temperature (in °C).
In this case, the mass of water is given as 1000.0g, and the initial and final temperatures are both 25.0°C. We want to find the heat gained by the water, so we'll use a positive value for q.
q_water = m_water * c_water * ΔT_water
q_water = 1000.0g * 4.184 J/g°C * (final temperature - 25.0°C)
Next, let's calculate the heat released by the dissolution of calcium chloride. The heat released by a reaction can be calculated using the equation:
q = ΔH_rxn
where:
q is the heat gained or lost (in joules), and
ΔH_rxn is the enthalpy change (in kilojoules) of the reaction.
In this case, ΔH_rxn for the dissolution of calcium chloride is given as +198.0 KJ. We want to find the heat released by the dissolution of calcium chloride, so we'll use a negative value for q.
q_CaCl2 = -198.0 KJ
Now, to calculate the final temperature of the solution, we need to set up an equation that balances the heat gained and heat lost:
q_water + q_CaCl2 = 0
Rearranging the equation, we have:
q_water = -q_CaCl2
Now substitute the expressions for q_water and q_CaCl2:
1000.0g * 4.184 J/g°C * (final temperature - 25.0°C) = -198.0 KJ
Simplify and solve for the final temperature:
1000.0g * 4.184 J/g°C * final temperature - 1000.0g * 4.184 J/g°C * 25.0°C = -198000 J
Now, solve for the final temperature:
1000.0g * 4.184 J/g°C * final temperature = -198000 J + 1000.0g * 4.184 J/g°C * 25.0°C
final temperature = (-198000 J + 1000.0g * 4.184 J/g°C * 25.0°C) / (1000.0g * 4.184 J/g°C)
Solving this equation will give you the final temperature of the solution.