Question

4. A 58.215 g sample of a pure metal element is brought to 99.0°C and added to 41.202 g of water at 01.5°C in a calorimeter. The metal and water arrive at a final, equal temperature of 27.2°C. a. Find the specific heat of the metal. b. Find the approximate atomic mass of the metal using the law of Dulong and Petit.

Answer 1

We can use the formula, Io = mx Cs XATI where = Heat m = mass Cs = specific heat AT = Temperature Given weight of sample = 58.215g weight of water = 41.202g Temp. of sample = 99°C Temp of water = 21.5°C final temperature = 27.2°C Note: The specific heat of water = 4.186J/g.c. • This is a thermo-equilibrins condition.. 2. Lees of heat or gain of heat by water] - Am =t&w] muutat x G x A Total = - Font Qw = 0 = ( m C G AT) metal + (m Cg AT) water 0 2) 58.215 x 68 x (27.2-99) + 41.202 * 4.18 ( 27.2-215) = (-4179.8 * (g) + 981.6=0 - Cs = -Q81 06/-4179.8 = 0.235J/g.c. - Cs & specific heat) = 0.2355/g.c)

@ Specific heat = 0.235J/g °C = 0.056 cal/g.oc - 6 the Duolong & Petit Law states that a product of specific heat of almost all solid elements and theire atomic mals is a i Atornic mass & specific heat = 6-4 Atonic maes - bilpecific heat - Atomic mais = 6.9 8.056 он Cadmium (cd) - Atomic mars = 110.7 | The metal sample wed might be lite
Formulas are used to solve the above questions. The values have beeen put and equated accordingly.

In the first question , there is a thermal equlibrium . So heat produced is zero. In the second question, the Duolong and Petit law is used as mentioned.