Question

What final temperature would you expect from a chemical hot pack with an initial temperature of 25.0 ^oC that mixes 20.00 g of CaCl₂(∆H = -82.9 kJ/mol) with 300.0 g water? Assume that the specific heat of the solution is 4.18 J/g^oC.

Answer 1

Given mass of CaCl2=20 g and mass of water=300 g.

Then total mass, m=20 g + 300 g=320 g.

Specific heat, C=4.18 J/g.°C,

Initial temperature, T1=25°C, final temperature, T2=?

Heat released,q=-82.9 kJ/mol=-82.9x10^3 J/mol.

Molar mass of CaCl2=110.98 g/mol.

Moles of CaCl2=mass/molar mass=20g/110.98 g/mol=0.18 mol.

For 1 mol of CaCl2, q=-8.9x10^3 J, then for 0.18 mol,

q=(0.18 mol x -82.9x10^3 J)/(1 mol)=-14939.628 J.

We know that heat, q=mc∆T=mC(T2-T1)

(-14939.628 J)=(320 g x 4.18 J/g.°C)(T2-25°C)

T2-25°C=-11.168 °C

T2=25-11.168 °C=13.83 °C.

Therefore final temperature is 13.83 °C.

Please let me know if you have any doubt. Thanks.