Question

1) If 100.0mL water of at 50.0C is added to 100.0 mL of water at 20.0C. What temperature would the mixture of water reach? Assume the system is perfectly sealed and no heat is lost to the calorimeter or to the outside environment.

2) If a student’s coffee cups are discarded after the first HCl reaction and new coffee cups are used only for the final reaction with acetic acid. How would this error impact the results?

3) A chemistry professor has a cup of coffee containing 50.0mL of room temperature coffee at 25.0C. The professor also has a new pot of hot coffee at temperature of 96.0C. What volume of hot coffee will the professor need to add to his cold coffee to reach the ideal drinking temperature of 82.0C? Assume that no heat is lost to the coffee cup or to the environment. Also assume that coffee has the same density (1.0g/mL) and heat capacity (4.184 J/g*C) as water.

***please answer all the question***

Answer 1

Ans. #1. Amount of gained or lost by water to attain thermal equilibrium is given by-

            q = m s dT                            - equation 1

Where,

q = heat change

m = amount of water in moles

s = specific heat of water [ 4.184 J g^-10C^-1]

dT = Final temperature – Initial temperature

# When a hot water sample is mixed with a cold water sample-

I. The hot water sample loses heat to the system (mixture). The heat is denoted by –ve value of q.

II. The cold water sample gains heat from the system (mixture).

            III. While attaining thermal equilibrium, the total heat lost by hot water must be equal to the heat gain by colt water.

# Let the temperature at thermal equilibrium = T⁰C

It’s assumed that the density of water is 1.000 g/mL. So, mass of water sample is equal to the respective numerical value of volume in grams.

Now,

            Heat lost by hot water = Heat gained by cold water

            Or, -q1 (hot water) = q2 (cold water)

            Or, - 100.0g x (4.184 J g^-10C^-1) x (T –50.0)⁰C = 100.0g x (4.184 J g^-10C^-1) x (T –20.0)⁰C

            Or, -(T – 50.0)⁰C = (T – 20.0)⁰C

            Or, 50 – T = T – 20

            Or, 2T = 50 + 20 = 70

            Hence, T = 35.0

Therefore, temperature at thermal equilibrium = T⁰C = 35.0⁰C

#2. It depends on the type of reaction taking place in the first calorimeter and temperature difference between the HCl solution and that of acetic acid taken.

Without the outline of experiment, it would be quite unwise to propose a solution.

However, if the temperature of HCl solution at thermal equilibrium and that of acetic acid differ, the calculated value of specific heat or any variable would differ.

#3. Taking the same approach as #1.

Let the mass of hot coffee = m

Now,

            Heat lost by hot coffee = Heat gained by cold water

            Or, -q1 (hot coffee) = q2 (cold coffee)

            Or, - m x (4.184 J g^-10C^-1) x (82-96)⁰C = 50.0g x (4.184 J g^-10C^-1) x (82 –25)⁰C

            Or, - m x (-14.0)⁰C = 50.0g x (57)⁰C

            Or, 14 m = 2850 g

            Or, m = 2850 g / 14 = 203.57 g

Hence, required mass of hot coffee = 203.57 g = 203.57 mL