In: Chemistry
1. An unknown metal is found to have a specific heat capactiy of 0.281. From this, calculate its molar mass.
1. A student measures the initial temperature of an object as 26.17oC and the final temperature as 56.82 oC. What is the change in temperature in units of Kelvin?
2 A student measures the initial temperature of an object as 26.17oC and the final temperature as 56.82 oC. What is the change in temperature in units of Kelvin?
3. 51.59 g of water at 82.6 oC is added to a calorimeter that contains 48.01 g of water at 40.9 oC. If the final temperature of the system is 59.4 oC, what is the calorimeter constant (Ccalorimeter) ? Use 4.184 J/goC for the heat capacity of water.
4.How much energy (in Joules) is required to heat 13.23 g of water from 16.5 oC to 49.9 oC ? Use 4.184 J/goC as the specific heat capacity of water.
Ans. #1. There is no direction correlation between the heat capacity of a substance and its molar mass. So, the molar mass of the metal can’t be calculated using “0.281” as its heat capacity.
#2. Change in temperature = Final temperature – Initial temperature
= 56.820C – 26.170C
= 30.650C
= 30.65 K
Note: The difference in 0C is equivalent to difference in K ; however temperature in 0C is NOT equal to temperature in K.
# Temperature in K = (0C + 273.15) K
So, 56.820C = (56.82 + 273.15) K
26.170C = (26.17 + 273.15) K
Now, change in temperature dT = (56.82 + 273.15) K - (26.17 + 273.15) K
= 56.82 K + 273.15 K – 26.17 K – 273.15 K
= 56.82 K – 26.17 K
= 30.65 K
Therefore, dT in 0C is equivalent to K.
#3. Amount of heat changed (gained or lost) during attaining thermal equilibrium is given by-
q = m s dT - equation 1
Where,
q = heat gained or lost
m = mass
s = specific heat
dT = Final temperature – Initial temperature
Temperature at thermal equilibrium = 59.40C
# Amount of heat lost by hot water, q1 = 51.59 g x (4.184 J g-10C-1) x (82.6 – 54.9)0C
Or, q1 = - 5979.115912 J
The –ve sign indicates the heat is being released as water cools.
# Amount of heat gained by water in calorimeter, q2 =
48.01 g x (4.184 J g-10C-1) x (54.9 – 40.9)0C
Or, q2 = 2812.23376 J
# At thermal equilibrium, total heat lost by hot water is equal to the sum of heat gained by cold water in calorimeter plus the calorimeter itself.
So,
Heat gained by calorimeter = Total heat lost by hot water – Heat gained by cold water
= 5979.115912 J - 2812.23376 J
= 3166.882152 J
Hence, heat gained by calorimeter = 3166.882152 J
Change in temperature of calorimeter while attaining thermal equilibrium =
Final temperature – Initial temperature
= 54.90C – 40.90C
= 14.00C
Now,
Calorimeter constant = Heat gained by calorimeter / Temperature change
= 3166.882152 J / 14.00C
= 226.21 J/ 0C
Therefore, calorimeter constant = 226.21 J/ 0C
#4. Putting the values in equation 1-
q = 13.23 g x (4.184 J g-10C-1) x (49.9 – 16.5)0C
Or, q = 1848.83 J
Hence, the amount of heat required = 1848.83 J