Question

Name Section Experiment 14 Data and Calculations: Heat Effects and Calorimetry A. Specific Heat Trial 1 Mass of stoppered test tube plus metal Mass of test tube and stopper Mass of calorimeter 7.2 g Mass of calorimeter and water 46-1 Mass of water 40-0 Mass of metal 113.6 21.2 . °c Initial temperature of water in calorimeter Initial temperature of metal (assume 100°C unless directed to do otherwise) 100 °c Equilibrium temperature of metal and water in calorimeter 25.1 3.9 Al water 'final - Sinicial - 74.9°C Atmental) 9H0 Specific heat of the metal (Eq. 3) J/gºC Approximate molar mass of metal Unknown metal # w

Answer 1

We know that, q_H2O = m₁c₁$\Delta$t_H2O

where, m₁ = mass of water = (46.1 - 7.2) g = 38.9 g,

c₁ = heat capacity of water = 4.186 J / g ^oC,

$\Delta$t_H2O = final - initial temperature = 3.9 ^oC

So, q_H2O = 38.9 g X 4.186 J / gC X 3.9 ^oC = 635.06 J

Now,

Heat gained by water = q_H2O = Heat lost by metal = q_metal = 635.06 J

$\Delta$t_metal = final - initial temperature = 74.9 ^oC

m₂ = 113.6 g

So, for metal: q_metal = m₂c₂$\Delta$t_metal$\Rightarrow$ c₂ = (635.06 J) / (113.6 g X 74.9 ^oC)

c₂ = 0.0746 J / g ^oC

The approximate value of molar mass can be calculated by using Dulong and Petit's law. It states that for a solid element the product of the specific heat capacity and its mass per mole is constant.

molar mass (g/mol) x specific heat (J/g.K) = 25 J/mol. K

Since in calculation of c₂ we have used change in temperature and 1 degree change in celsius scale is similar to 1 degree change in kelvin scale. So, c₂ = 0.0746 J / g ^oC = 0.0746 J / gK

$\Rightarrow$ molar mass = (25 J/mol. K) / 0.0746 J / gK = 335.12 g/mol

Hope this helps!