Need Chem help how do I calculate blanks Mass of empty calorimeter (g) Mass of c

Question

Need Chem help how do I calculate blanks Mass of empty calorimeter (g) Mass of calorimeter and water (g) Mass of water (g) Mass of empty test tube (g) Water temperature after the addition of hot lead. Mass of test tube and lead (g) Time Temp Mass of lead (g) Initial temperature of water Temperature of hot lead (degree C) Final temperature of water degree C) Change in temperature of water (degree C) Heat added to water (cal) Change in temperature of lead (degree C) Experimental specific heat (cal/g degree C) Specific heat from table (cal/8 degree C) Percent Error (%)

Explanation / Answer

To calculate specific heat of Lead

q(water) = mCpdT

with,

mass of water (m) = 50 g

Cp = specific heat of water

change in temperature of water (dT) = 32.8 - 29 = 3.8 oC

we get,

heat added to water (q) = 50 x 4.184 x 3.8 = 794.96 J = 190 cal

Then,

heat lost by lead = heat added to water

so,

q(lead) = mCpdT = 190 cal

mass of lead = 30 g

Cp = specific heat of metal

change in temperature of lead (dT) = 91 - 32.8 = 58.2 oC

So,

q(lead) = 190 = 30 x Cp x 58.2

Experimental specific heat Cp(lead) = 0.109 cal/g.oC

Specific heat from table = 0.0306 cal/g.oC

Percent error = (0.109 - 0.0306) x 100/0.109 = 71.93%

[please note : a very large variation is seen here between the literature reported speicific heat and the experimental value. The error is huge.]

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