382 Report Sheet Heat of Neutralization A piece of metal weighing 5.10 g at a te

Question

382 Report Sheet Heat of Neutralization A piece of metal weighing 5.10 g at a temperature of 48.6°C was placed in a calorimeter in 20.00 mL of water at 22.1°C·The final equilibrium temperature was found to be 292°C. What is the specific heat of the metal? 8, 9. If the specific heat of methanol is 2.51 J/K-g, how many joules are necessary to raise the temperature of 50 g of methanol from 20°C to 70°C? 10. When a 3.25 g sample of solid sodium hydroxide was dissolved in a calorimeter in 100.0 g of water, the temperature rose from 23.9 to 32.0. Calculate /l (in kl mol NaOH) for the following solution process: NaOH(s)Na (ag)+OH(a) Assume that it's a perfect calorimeter and that the specific heat of the solution is the same as that of pure water.

Explanation / Answer

Q = mcT

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kgK), is a symbol meaning "the change in"

T = change in temperature (Kelvins, K)

Question 8

Heat lost by metal = Heat gained by water

Mass of metal = 5.1 gm

temperature of metal = 48.6 0C.

Mass of water = 20 ml = 20 gm

Temperature of water = 22.1 0C.

Specific heat of water = 4.184 J/g0C

5.1 x sp ht x (48.6-29.2) = 20 x 4.184 x (29.2-22.1)

sp ht of Metal = 6.00  J/g0C

Question 9

Mass of methanol = 50 gm

T = 70 -20 = 500C

Specific heat of methanol = 2.51 J / K g

Q = 50 x 50 x 2.51 = 6275 Joules

6275 Joules of heat is needed

Question 10

Mass of water and NaOH = 100 +3.25 = 103.25 gm

T =  32-23.9 = 8.1 0C

Specific heat of water = 4.184 J / 0C g

Heat = 103.25 x 8.1 x 4.184 = 3499.18 Joules

Moles of NaOH = 3.25 x 1000 / 40 x 103.25 = 0.7869 M

Delta H of NaOH = 3499.18 / 0.7869 = 4446.79 Joules

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