1. An unknown metal is found to have a specific heat capactiy of 0.281. From thi

Question

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.

Explanation / Answer

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

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