The temperature of 500 mL of distilled water in a flask is between 70°C and 90°C

Question

The temperature of 500 mL of distilled water in a flask is between 70°C and 90°C. Room temperature is 20°C. the useful range of the only thermometer available is between 0°C and 50°C. Briefly describe an experiment you could carry out which would allow you to determine exactly the temperature of the water in the flask. Indicate the following in your description:

What equipment would you use?

What would you do?

The equation(s) for any calculation(s) you would carry out, and which quantities would be measured, which would be calculated, and which would be constants.

Explanation / Answer

Ans. Instrument used is Coffee cup calorimeter.

Steps:

1. Place 100.0 g water at 200C in the calorimeter.

2. Transfer 20.0 g water of unknown temperature (T0C) to it.

3. Allow the system to attain thermal equilibrium. Note the equilibrium temperature. Say, the equilibrium temperature is 300C. [Note: taking a known value, as you know equilibrium temperature is useful to get a clear view.]

4. The water at room temperature gains heat that is lost be hot water (T0C) while attaining thermal equilibrium.

At thermal equilibrium,

            Heat gained by water at 200C = heat lost by water at T0C

Now, Using,       q = m c dT       -- equation 1   

Where, q = heat change                                                

m= mass in gram

c = specific heat (in terms of J/g0C) = 4.18 J/ g0C            

dT = Equilibrium temperature – Initial temperature

Putting the respective values in equation 1-

            Heat gained by water at 200C = heat lost by water at T0C

            100 g x (4.18 J/ g0C) x (300C – 200C) = 20 g x (4.18 J/ g0C) x (T0C – 300C)

Or, 500C = (T0C – 300C)

Or, T0C = 500C + 300C = 800C

Thus, the unknown temperature calculated to be 800C.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.