A. Specific Heat Trial 1 Trial 2 42.900 y2.810g 2I. I05 1.301 41930 Mass of stop
ID: 537983 • Letter: A
Question
A. Specific Heat Trial 1 Trial 2 42.900 y2.810g 2I. I05 1.301 41930 Mass of stoppered test tube plus metal of ppened test the plas metal Mass of test tube and stopper 7.304 g Mass of calorimeter 2.513 Mass of calorimeter and watct 40.I7 21.11 54.51 - 21.195 Mass of water 21.77i 23,2 F21.745 Mass of metal lutal temperature of water in cal rameter Initial temperature of water in calorimeter Eoc Initial temperature of mctal (assume 100C unless ditected to do otherwise) i0o Eqguilibrium temperature of metal and water n calorimctcr 25-it 75.4c 74.9 320 S )OIec 0.252c C) 4a0 Specific heat of the mctal (Eç. 3 Approximate molar mass of metal 130 383 (0.2s Unknown metal # B. Heat of Solution Mass of calorimeter plus water Mass of beaker 2 C3.433, 2220 (coetinued on following page)Explanation / Answer
For Trial -1
Specific heat of metal can be calculated by comparing it with water.
Q = m c (tfti)
q = heat energy in joules
c = the specific with units joules/(grams x Celsius)
tf = final temperature and
ti = initial temperature.
The q for the metal is negative because it loses heat
The q for the water is positive because it absorbs heat
Since the system will reach thermal equilibrium we will make these equal
- q metal = q water
-(21.176g)(x)(25.1 C- 100.0 C)= (40.176 g)(4.184 J/gC)(25.1 C – 23.2 C)
1581.08 x = 319.38
x = 0.201 J/g 0 C
Specific heat of metal = 0.201 J/g 0 C
For Trial -2- q metal = q water
-(21.795g)(x)(25.4 C- 100.0 C)= (54.514 g)(4.184 J/gC)(25.4 C – 23.6 C)
1625.9 x = 410 56
x = 0.253 J/g 0 C
Specific heat of metal = 0.253 J/g 0 C
Heat of Solution-
In this experiment dissolve salt absorbs heat energy from the surrounding water, which causes the temperature of water to go down from 23.0 0C to 17.8 0C.
Use the formula Q = mass of water x specific heat of water x change in Temperature to calculate amount of heat gained by water.
q = 55.680 g x 4.18 J / g 0C x - 5.2 0C
q = - 1210.26 J = - 1.21 KJ
Change in enthalpy = Q gained by salt
H = 1.21 KJ
Heat of Solution/g =- 1210.26 J/5.02g = -241.1J/g
H = 241.1J/g