Question

A 1.00 g sample of coffee beans was completely combusted in a bomb calorimeter (Ccal = 62.1 J/o C) and caused the temperature of the water in the calorimeter to increase from 24.66 K to 27.22 K. What is the change in temperature of the water in degrees Celsius? The heat released by the coffee beans was: ? One cup of black coffee is made from 237 g of coffee beans. How many Calories are in one cup of black coffee (ignoring any other additives)? Which of the following is a correct relationship? A. ΔH = qp and ΔE = qv B. w > 0 at constant volume C. ΔH = qv and ΔE = qp D. ΔH > ΔE for all chemical processes?

Answer 1

Ans. #1. Change in temperature, dT = Final temperature – Initial temperature

                                                = 27.22 K – 24.66 K

                                                = 2.56 K

# Temperature in K = (K – 273.15)⁰C

So,

            dT = (27.22 – 273.15)⁰C - ( 24.66– 273.15)⁰C

            Or, dT = 27.22⁰C – 273.15⁰C -24.66⁰C + 273.15⁰C

            Or, dT = 2.56⁰C

Note that difference in kelvin and ⁰C unit is the same.

# Heat released by coffee beans, Q = C x dT

                        Where, C = heat capacity of the calorimeter

            Or, Q = (62.1 J ⁰C^-1) x 2.56⁰C

            Hence, Q = 158.976 J

#2. From #1. 1.0 g coffee beans release 158.976 J energy. So, calorific value of coffee beans = 158.976 J/ g

Now,

Energy in 1 cup of black coffee = Calorific value x Amount of coffee beans in one cup

                                                = (158.976 J/ g) x 237.0 g

                                                = 37677.312 J                                   ; [1 cal = 4.184 J]

                                                = 9005.094 cal        

#3.

# Option A. Incorrect                      

dH = qp                     -at constant pressure.

Since the pressure increase due to releases of CO2 and H2O during combustion, the pressure does not remains constant. But, there is increase in pressure of the system.

dE = qv                      - at constant volume

Option B. Incorrect.

w = p dV                    - at constant volume, dV = 0; So, w = p x 0 = 0

Since the volume of calorimeter remains constant, change in volume dV = 0.

Therefore, work done, w = 0

# Option C. Incorrect.

            dH = q x C                ; C = heat capacity.

# Option D. Incorrect. For reactions that result in a net production of gas,

dE < dH          - when, there is net production of gases (as during combustion in this case)

dE > dH          - when, there is net consumption of gases

Result: None of the options are correct.