What Is the Final Temperature of a Squeezed Hot Pack with LiCl and Water?

[Not a Wrench]

Homework Statement

What is the final temperature in a squeezed hot pack that contains 23.9 g of LiCl dissolved in 107 mL of water? Assume a specific heat of 4.18 J/(g⋅∘C) for the solution, an initial temperature of 25.0 ∘C, and no heat transfer between the hot pack and the environment.

Homework Equations

q=mcΔT
LiCl(s)⟶Li+(aq)+Cl−(aq)ΔH=−36.9kJ

The Attempt at a Solution

1. Change 23.9 g LiCl to mol. I get .564 mol LiCl
2. Use .564 mol LiCl x -36.9kJ to get -20.8 kJ for this reaction
3. Add masses of water and LiCl to get 130.9 g total.
4. Change -20.8 kJ to J: -20800 J
5. use q=mcΔT to get -20800J=(130.9g)(4.18g⋅°C)(Tf-25.0°C)
6. Find Tf to be -13.0°C.
7. Answer is wrong and I am confused. Please could someone help out?

 

[Chestermiller]

Homework Statement

What is the final temperature in a squeezed hot pack that contains 23.9 g of LiCl dissolved in 107 mL of water? Assume a specific heat of 4.18 J/(g⋅∘C) for the solution, an initial temperature of 25.0 ∘C, and no heat transfer between the hot pack and the environment.

Homework Equations

q=mcΔT
LiCl(s)⟶Li+(aq)+Cl−(aq)ΔH=−36.9kJ

The Attempt at a Solution

1. Change 23.9 g LiCl to mol. I get .564 mol LiCl
2. Use .564 mol LiCl x -36.9kJ to get -20.8 kJ for this reaction
3. Add masses of water and LiCl to get 130.9 g total.
4. Change -20.8 kJ to J: -20800 J
5. use q=mcΔT to get -20800J=(130.9g)(4.18g⋅°C)(Tf-25.0°C)
6. Find Tf to be -13.0°C.
7. Answer is wrong and I am confused. Please could someone help out?

The sum of the enthalpy change for the reaction PLUS the sensible change in enthalpy of the products must be equal to zero.

 

[Not a Wrench]

The sum of the enthalpy change for the reaction PLUS the sensible change in enthalpy of the products must be equal to zero.

I don't know what that means, really.

 

[Chestermiller]

I don't know what that means, really.

It means $$-20800J+(130.9g)(4.18\frac{J}{g\ C})(Tf-25.0°C)=0$$

 

[Not a Wrench]

It means $$-20800J+(130.9g)(4.18\frac{J}{g\ C})(Tf-25.0°C)=0$$

I never learned this during my lecture. Could you explain why this is?

 

[Borek]

It is just another way of writing energy conservation in a closed system.

 

[Not a Wrench]

It is just another way of writing energy conservation in a closed system.

Oh, it's conservation of ΔE? Now I see.

 

[Chestermiller]

I never learned this during my lecture. Could you explain why this is?

There are two ways of looking at it, and both ways give the same answer.

Way 1: The process takes place in a closed system at constant pressure. So, from the first law of thermodynamics, $Q = \Delta H$. But the process is adiabatic, so the overall change in enthalpy is equal to zero: $\Delta H=0$. The change in enthalpy is comprised of 2 parts: the heat of reaction (negative for exothermic and positive for endothermic) plus the increase in sensible heat (enthalpy) of the products. This leads to the equation that I wrote: $$\Delta H_R+nC_p\Delta T=0$$

Way 2: The heat of reaction $\Delta H_R$ is defined as the amount of heat that would have to be added to the system in order for the final temperature of the products to be equal to the initial temperature of the reactants. For an exothermic reaction, you would have to remove heat to do this, so $\Delta H_R$ would be negative; for an endothermic reaction, you would have to add heat to do this, so $\Delta H_R$ would be positive. But, this system is adiabatic, so the temperature of the products will have to be different than the temperature of the reactants. If the reaction were endothermic, we would have to remove the heat of reaction from the products in order for no net heat to have entered the system. If the reacton were exothermic, we would have to add back in the heat of reaction to the products in order for no net heat to have left the system. In either case, we would have to write that: $$nC_p\Delta T = -\Delta H_R$$