What is the final temperature in a squeezed hot pack that contains 27.6 g of LiCl dissolved in 102 mL of water?

asked by @sarahe44 • almost 2 years ago • Chemistry • 5 pts

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.

Instant hot packs contain a solid and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, increasing the temperature because of the exothermic reaction. The following reaction is used to make a hot pack: LiCl(s)⟶Li+(aq)+Cl−(aq)ΔH=−36.9kJ

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1 answer

Hey! For this problem, we will be using the formula Q=MCdeltaT.

The Q represents energy, which is given to you at the end of the problem by the enthalpy or deltaH (it is negative since this is an exothermic reaction). Although we are given 36.9 kJ, we want to convert that to Joules, since our C (specific heat capacity) is in terms of Joules. So 36.9 kJ x (1000 J/ 1 kJ) = 36,900 J. This is your Q value now.

Next, M represents the mass. Here, the mass is 27.6 g of LiCl. C represents your specific heat capacity which is given to you in the problem = 4.18 J/ g. C. Finally, delta T stands for the difference in temperatures ( Tf-Ti). You are given your initial temperature (Ti) = 25 C. We are solving for the Tf or final temperature. So we plug in (Tf-Ti) in the Q=MC(Tf-Ti). Now we isolate Tf---> Tf = (Q/MC) + Ti. Plug in all of the values and you should obtain a final temperature or Tf = 344.84 C.

Hope that helped!

answered by @sabrina • almost 2 years ago
  • Why do you change the Q from negative to positive in the calculation? Alexa commented 11 months ago